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## How to Easily Solve Trigonometric Equations

Last Updated: February 1, 2023 Fact Checked

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 17 people, some anonymous, worked to edit and improve it over time. There are 10 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 256,909 times. Learn more...

Did you get homework from your teacher that was about solving Trigonometric equations? Did you maybe not pay full attention in class during the lesson on Trigonometric questions? Do you even know what "Trigonometric" means? If you answered yes to these questions, then you don't need to worry because this wikiHow will teach you how to solve Trigonometric equations.

• To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations.

• There are 4 types of basic trig equations:
• sin x = a ; cos x = a
• tan x = a ; cot x = a
• Solving basic trig equations proceeds by studying the various positions of the arc x on the trig circle, and by using trig conversion table (or calculator ). To fully know how to solve these basic trig equations, and similar, see book titled :"Trigonometry: Solving trig equations and inequalities" (Amazon E-book 2010).
• Example 1. Solve sin x = 0.866. The conversion table (or calculator) gives the answer: x = Pi/3. The trig circle gives another arc (2Pi/3) that has the same sin value (0.866). The trig circle also gives an infinity of answers that are called extended answers.
• x1 = Pi/3 + 2k.Pi, and x2 = 2Pi/3. (Answers within period (0, 2Pi))
• x1 = Pi/3 + 2k Pi, and x2 = 2Pi/3 + 2k Pi. (Extended answers).
• Example 2. Solve: cos x = -1/2. Calculators give x = 2 Pi/3. The trig circle gives another x = -2Pi/3.
• x1 = 2Pi/3 + 2k.Pi, and x2 = - 2Pi/3. (Answers within period (0, 2Pi))
• x1 = 2Pi/3 + 2k Pi, and x2 = -2Pi/3 + 2k.Pi. (Extended answers)
• Example 3. Solve: tan (x - Pi/4) = 0.
• x = Pi/4 ; (Answer)
• x = Pi/4 + k Pi; ( Extended answer)
• Example 4. Solve cot 2x = 1.732. Calculators and the trig circle give
• x = Pi/12 ; (Answer)
• x = Pi/12 + k Pi ; (Extended answers)

• To transform a given trig equation into basic trig ones, use common algebraic transformations ( factoring , common factor , polynomial identities...), definitions and properties of trig functions, and trig identities. There are about 31, among them the last 14 trig identities, from 19 to 31, are called Transformation Identities, since they are used in the transformation of trig equations. [4] X Research source See book mentioned above.
• Example 5: The trig equation: sin x + sin 2x + sin 3x = 0 can be transformed, using trig identities, into a product of basic trig equations: 4cos x*sin (3x/2)*cos (x/2) = 0. The basic trig equations to be solved are: cos x = 0 ; sin (3x/2) = 0 ; and cos (x/2) = 0.

• Before learning solving trig equations, you must know how to quickly find the arcs whose trig functions are known. Conversion values of arcs (or angles) are given by trig tables or calculators. [6] X Research source
• Example: After solving, get cos x = 0.732. Calculators give the solution arc x = 42.95 degree. The trig unit circle will give other solution arcs that have the same cos value.

• You can graph to illustrate the solution arcs on the trig unit circle. The terminal points of these solution arcs constitute regular polygons on the trig circle. For examples:
• The terminal points of the solution arcs x = Pi/3 + k.Pi/2 constitute a square on the trig unit circle.
• The solution arcs x = Pi/4 + k.Pi/3 are represented by the vertexes of a regular hexagon on the trig unit circle.

• A. Approach 1.
• Transform the given trig equation into a product in the form: f(x).g(x) = 0 or f(x).g(x).h(x) = 0, in which f(x), g(x) and h(x) are basic trig equations.
• Example 6. Solve: 2cos x + sin 2x = 0. (0 < x < 2Pi)
• Solution. Replace in the equation sin 2x by using the identity: sin 2x = 2*sin x*cos x.
• cos x + 2*sin x*cos x = 2cos x*( sin x + 1) = 0. Next, solve the 2 basic trig functions: cos x = 0, and (sin x + 1) = 0.
• Example 7. Solve: cos x + cos 2x + cos 3x = 0. (0 < x < 2Pi)
• Solution: Transform it to a product, using trig identities: cos 2x(2cos x + 1 ) = 0. Next, solve the 2 basic trig equations: cos 2x = 0, and (2cos x + 1) = 0.
• Example 8. Solve: sin x - sin 3x = cos 2x. (0 < x < 2Pi)
• B. Approach 2.
• Transform the given trig equation into a trig equation having only one unique trig function as variable. There are a few tips on how to select the appropriate variable. The common variables to select are: sin x = t; cos x = t; cos 2x = t, tan x = t and tan (x/2) = t.
• Example 9. Solve: 3sin^2 x - 2cos^2 x = 4sin x + 7 (0 < x < 2Pi).
• Solution. Replace in the equation (cos^2 x) by (1 - sin^2 x), then simplify the equation:
• 3sin^2 x - 2 + 2sin^2 x - 4sin x - 7 = 0. Call sin x = t. The equation becomes: 5t^2 - 4t - 9 = 0. This is a quadratic equation that has 2 real roots: t1 = -1 and t2 = 9/5. The second t2 is rejected since > 1. Next, solve: t = sin = -1 --> x = 3Pi/2.
• Example 10. Solve: tan x + 2 tan^2 x = cot x + 2.
• Solution. Call tan x = t. Transform the given equation into an equation with t as variable: (2t + 1)(t^2 - 1) = 0. Solve for t from this product, then solve the basic trig equation tan x = t for x.

• There are a few special types of trig equations that require some specific transformations. Examples:
• a*sin x+ b*cos x = c ; a(sin x + cos x) + b*cos x*sin x = c ;
• a*sin^2 x + b*sin x*cos x + c*cos^2 x = 0

• The function f(x) = sin x has 2Pi as period.
• The function f(x) = tan x has Pi as period.
• The function f(x) = sin 2x has Pi as period.
• The function f(x) = cos (x/2) has 4Pi as period.
• If the period is specified in the problem/test, you have to only find the solution arc(s) x within this period.
• NOTE: Solving trig equation is a tricky work that often leads to errors and mistakes. Therefore, answers should be carefully checked. After solving, you can check the answers by using a graphing calculator to directly graph the given trig equation R(x) = 0. The answers (real roots) will be given in decimals. For example, Pi is given by the value 3.14
• ↑ https://www.purplemath.com/modules/solvtrig.htm
• ↑ https://www.mathopenref.com/arcsin.html
• ↑ https://courses.lumenlearning.com/precalculus/chapter/unit-circle-sine-and-cosine-functions/
• ↑ https://mathbitsnotebook.com/Algebra2/TrigConcepts/TCEquationsMore.html
• ↑ https://www.analyzemath.com/trigonometry/properties.html

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## Trigonometric Equations

The trigonometric equations involve trigonometric functions of angles as variables. The angle of θ trigonometric functions such as Sinθ, Cosθ, Tanθ is used as a variable in trigonometric equations. Similar to general polynomial equations, the trigonometric equations also have solutions, which are referred to as principal solutions, and general solutions.

We will use the fact that the period of sin x and cos x is 2π and the period of tan x is π to find the solutions of the trigonometric equations. Let us learn more about trigonometric equations, the method to solve them, and find their solutions with the help of a few solved examples of trigonometric equations for a better understanding of the concept.

## What are Trigonometric Equations?

The trigonometric equations are similar to algebraic equations and can be linear equations, quadratic equations, or polynomial equations. In trigonometric equations, the trigonometric ratios of Sinθ, Cosθ, Tanθ are represented in place of the variables, as in a normal polynomial equation. The trigonometric ratios used in trigonometric equations are Sinθ, Cosθ, or Tanθ.

The linear equation ax + b = 0 can be written as a trigonometry equation as aSinθ + b = 0, which is also sometimes written as Sinθ = Sinα. The quadratic equation ax 2 + bx + c = 0 is as an example of trigonometric equation is written as aCos 2 θ + bCosθ + c = 0. But unlike normal solutions of equations with the number of solutions based on the degree of the variable, in trigonometric equations, the same value of solution exists for different values of θ. For example, we have Sinθ = 1/2 = Sinπ/6 = Sin5π/6 = Sin13π/6, and so on as the values of the sine function repeat after every 2π radians.

Some of the examples of trigonometric equations are as follows.

• Sin2x - Sin4x + Sin6x = 0
• 2Cos 2 x + 3Sinx = 0
• Cos4x = Cos2x
• Sin2x + Cosx = 0
• Sec 2 2x = 1 - Tan2x

## Trigonometric Equations Formulas

We use some results and general solutions of the basic trigonometric equations to solve other trigonometric equations. These results are as follows:

• For any real numbers x and y, sin x = sin y implies x = nπ + (-1) n y, where n ∈ Z.
• For any real numbers x and y, cos x = cos y implies x = 2nπ ± y, where n ∈ Z.
• If x and y are not odd multiples of π/2, then tan x = tan y implies x = nπ + y, where n ∈ Z.

Now, we can prove these results using trigonometric formulas.

## Prove that for any real numbers x and y, sin x = sin y implies x = nπ + (-1) n y, where n ∈ Z

Proof: If sin x = sin y, then sin x – sin y = 0

⇒ 2 cos (x + y)/2 sin (x − y)/2 = 0 --- [Using formula Sin A - Sin B = 2 cos ½ (A + B) sin ½ (A - B) ]

⇒ cos (x + y)/2 = 0 or sin (x − y)/2 = 0

⇒ (x + y)/2 = (2n + 1)π /2 or (x − y)/2 = nπ, where n ∈ Z ---- [Because sin A = 0 implies A = nπ and cos A = 0 implies A = (2n + 1)π/2, where n ∈ Z]

i.e. x = (2n + 1) π – y or x = 2nπ + y, where n ∈ Z.

Hence x = (2n + 1)π + (–1) 2n + 1 y or x = 2nπ + (–1) 2n y, where n ∈ Z.

Combining these two results, we get x = nπ + (–1) n y, where n ∈ Z.

## Prove that for any real numbers x and y, cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

Proof: If cos x = cos y, then cos x – cos y = 0

⇒ -2 sin (x + y)/2 sin (x − y)/2 = 0 --- [Using formula Cos A - Cos B = - 2 sin ½ (A + B) sin ½ (A - B) ]

⇒ sin (x + y)/2 = 0 or sin (x − y)/2 = 0

⇒ (x + y)/2 = nπ or (x − y)/2 = nπ, where n ∈ Z ---- [Because sin A = 0 implies A = nπ, where n ∈ Z]

i.e. x = 2nπ – y or x = 2nπ + y, where n ∈ Z.

Hence x = 2nπ ± y, where n ∈ Z.

## Prove that if x and y are not odd multiples of π/2, then tan x = tan y implies x = nπ + y, where n ∈ Z.

Proof: If tan x = tan y, then tan x - tan y = 0

⇒ sin x / cos x - sin y / cos y = 0

⇒ (sin x cos y - cos x sin y) / (cos x cos y) = 0

⇒ sin (x - y) / (cos x cos y) = 0 ---- [Using trigonometric formula sin (A - B) = sinA cosB - sinB cosA ]

⇒ sin (x - y) = 0

⇒ x - y = nπ, where n ∈ Z --- [Because sin A = 0 implies A = nπ, where n ∈ Z]

⇒ x = nπ + y, where n ∈ Z

## Solving Trigonometric Equations

Unlike normal solutions of algebraic equations with the number of solutions based on the degree of the variable, in trigonometric equations, the solutions are of two types, based on the different value of angle for the trigonometric function , for the same solution. For example, for a simple trigonometric equation 2Cosθ - 1 = 0, the solution is given by, Cosθ = 1/2 and, the θ values are π/3, 5π/3, 7π/3, 11π/3, and so on as the values of the cosine function repeat after every 2π radians and cos x is positive in the first and fourth quadrants. We have two types of solutions to the trigonometric equations:

• Principal Solution: The initial values of angles for the trigonometric functions are referred to as principal solutions. The solution of Sinx and Cosx repeat after an interval of 2π, and the solution of Tanx repeat after an interval of π. The solutions of these trigonometric equations, for which x lies between 0 and 2π, are called principal solutions.
• Sinθ = Sinα, and the general solution is θ = nπ + (-1) n α, where n ∈ Z
• Cosθ = Cosα, and the general solution is θ = 2nπ + α, where n ∈ Z
• Tanθ = Tanα, and the general solution is θ = nπ + α, where n ∈ Z

## Steps to Solve Trigonometric Equations

The following steps are to be followed, for solving a trigonometric equation.

• Transform the given trigonometric equation into an equation with a single trigonometric ratio (sin, cos, tan)
• Change the equation with the trigonometric equation, having multiple angles, or submultiple angles into a simple angle.
• Now represent the equation as a polynomial equation , quadratic equation, or linear equation.
• Solve the trigonometric equation similar to normal equations, and find the value of the trigonometric ratio.
• The angle of the trigonometric ratio or the value of the trigonometric ratio represents the solution of the trigonometric equation.

## Examples of Solving Trigonometric Equations

Example 1: Find the principal solutions of the trigonometric equation sin x = √3/2.

Solution: To find the principal solutions of sin x = √3/2, we know that sin π/3 = √3/2 and sin (π - π/3) = √3/2

⇒ sin π/3 = sin 2π/3 = √3/2

We can find other values of x such that sin x = √3/2, but we need to find only those values of x such that x lies in [0, 2π] because a principal solution lies between 0 and 2π.

So, the principal solutions of sin x = √3/2 are x = π/3 and 2π/3.

Example 2: Find the solution of cos x = 1/2.

Solution: In this case, we will find the general solution of cos x = 1/2. We know that cos π/3 = 1/2, so we have

cos x = 1/2

⇒ cos x = cos π/3

⇒ x = 2nπ + (π/3), where n ∈ Z ---- [Using Cosθ = Cosα, and the general solution is θ = 2nπ + α, where n ∈ Z]

Therefore, the general solution of cos x = 1/2 is x = 2nπ + (π/3), where n ∈ Z.

Important Notes on Trigonometric Equations

• sin A = 0 implies A = nπ and cos A = 0 implies A = (2n + 1)π/2, where n ∈ Z

Related Topics

• Trigonometric functions
• Trigonometric Chart
• Trigonometric Identities
• Inverse Trigonometric Ratios

## Trigonometric Equations Examples

Example 1: Find the principal solution of the trigonometric equation tan x = -√3

Solution: Here we have tan x = -√3, and we know that tan π/3 = √3. So, we have

tan x = -√3

⇒ tan x = - tan π/3

⇒ tan x = tan(π - π/3) OR tan x = tan(2π - π/3)

⇒ tan x = tan 2π/3 OR tan x = tan 5π/3.

Therefore, the principal solutions of tan x = -√3 are 2π/3 and 5π/3

Answer: Principal solutions are x = 2π/3 and x = 5π/3.

Example 2: Find the general solution of the trigonometric equation 2 cos 2 x + 3 sin x = 0.

Solution: The given trigonometric equation is 2 cos 2 x + 3 sin x = 0.

⇒ 2(1 - sin 2 x) + 3 sin x = 0 --- [Using trigonometric formula cos 2 x + sin 2 x = 1 which implies cos 2 x = 1 - sin 2 x]

⇒ 2 - 2 sin 2 x + 3 sin x = 0

⇒ 2 sin 2 x - 3 sin x - 2 = 0

⇒ 2sin 2 x - 4 sin x + sin x - 2 = 0

⇒ 2 sin x(sin x - 2) + 1(sin x - 2) = 0

⇒ (2sin x + 1)(sin x - 2) = 0

⇒ 2 sin x + 1 = 0, sin x - 2 = 0

⇒ sin x = -1/2 or sin x = 2

⇒ sin x = 2 is not defined since the values of sin x lies between -1 and +1.

So, we have sin x = -1/2. We know that sin π/6 = 1/2.

sin x = -1/2

⇒ sin x = - sin (π/6)

⇒ sin x = sin (-π/6)

= sin (π + π/6)

= sin (7π/6)

Therefore, we have x = nπ + (-1) n (7π/6), where n ∈ Z --- [Because for any real numbers x and y, sin x = sin y implies x = nπ + (-1) n y, where n ∈ Z.]

Answer: General solution of the above trigonometric equation is nπ +(-1) n (7π/6), where n ∈ Z

Example 3: Solve sin 2x – sin 4x + sin 6x = 0.

Solution: The equation sin 6x + sin 2x − sin 4x = 0 can be written as

2 sin 4x cos2x − sin 4x = 0 --- [Using sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2]

⇒ sin 4x(2 cos2x − 1) = 0

⇒ sin 4x = 0 or 2 cos2x - 1 = 0

⇒ 4x = nπ or cos 2x = 1/2, , where n ∈ Z

⇒ 4x = nπ or cos 2x = cos π/3 --- [Because cos π/3 = 1/2], , where n ∈ Z

⇒ x = nπ/4 or 2x = 2nπ ± (π/3), , where n ∈ Z

⇒ x = nπ/4 or x = nπ ± (π/6), , where n ∈ Z

Answer: The generla solution is x = nπ/4 or x = nπ ± (π/6).

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## Trigonometric Equations Questions

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## FAQs on Trigonometric Equations

The trigonometric equations are similar to algebraic equations and can be linear equations, quadratic equations, or polynomial equations. In trigonometric equations, the trigonometric ratios of Sinθ, Cosθ, Tanθ are represented in place of the variables.

## What are the Three Trigonometric Equations Formulas?

The three trigonometric equations are based on the three trigonometric functions . The three trigonometric equations are Sinθ = Sinα, Cosθ = Cosα, Tanθ = Tanα. The three trigonometric equations formulas are as follows:

## Give Trigonometric Equations Examples.

Some of the examples of trigonometric equations are as follows:

## What are the Steps to Solve Trigonometric Equations?

The following steps are helpful for solving a trigonometric equation:

• Transform the given trigonometric equation into an equation with a single trigonometric ratio.
• Change the equation with the trigonometric equation, having multiple angles, or submultiple angles into a simple. angle.
• Now represent the equation as a polynomial equation, quadratic equation, or linear equation.

## What is the Difference Between Trigonometric Equations and Algebraic Equations?

The trigonometric equation and the algebraic equations differ in the variable used for the equation. In a trigonometric equation, the trigonometric function is the variable, and in an algebra the alphabets x, y are taken as variables. Both the trigonometric equations and algebraic equations have linear equations, quadratic equations, and polynomial equations.

## What is the Difference Between Trigonometric Equations and Identities?

Trigonometric Equations are like algebraic equations where we are required to determine the solution and find the value of the variable. On the other hand, trigonometric identities are results in trigonometry that express the relationship between different trigonometric ratios.

## What is the Principal Solution of Trigonometric Equations?

we know that sin x and cos x repeat after every 2π radians and tan x repeat after every π radians. The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions.

## How to Find the General Solution of Trigonometric Equations?

We can find the general solution of trigonometric equations using the following three results:

• For any real numbers x and y, sin x = sin y, implies x = nπ + (-1) n y, where n ∈ Z.
• For any real numbers x and y, cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

## How to Solve Trigonometric Equations

Summary: A trigonometric equation is one that involves one or more of the six functions sine, cosine, tangent, cotangent, secant, and cosecant. Some trigonometric equations, like x  = cos  x , can be solved only numerically, through successive approximations. But a great many can be solved analytically — in “closed form”, an exact solution in symbols — and this page shows you how to do it in five steps.

## Step 1. Get one function of one angle.

Step 2. solve for the value(s) of a trig function., step 3. solve for the angle., step 4. solve for the variable., step 5. apply any restrictions., more examples.

If a trig equation can be solved analytically, these steps will do it:

On this page I’ll walk you through solving several trig equations using these steps, showing you every detail. Once you know and understand the steps, you’ll be able to work some more examples more quickly.

Trig equations, like any equations, are really about numbers, not angles. You are looking for all possible numbers that could be substituted for the variable in the equation to make it true. But it simplifies things to think about the angles first and worry about the variables later.

Example A :

cos(4 A ) − sin(2 A ) = 0

Here the “angles”, the arguments to the trig functions, are 4 A and 2 A . True, you want to solve for A ultimately. But if you can solve for the angle 4 A or 2 A , it is then quite easy to solve for the variable .

As you see, that equation involves two functions (sine and cosine) of two angles (4 A and 2 A ). You need to get it in terms of one function of one angle. Note well: a function of one angle, not necessarily a function of just the variable A .

This is where it is essential to have a nodding acquaintance with all the trig identities . If you do, you’ll remember that cos(2 u ) can be expressed in terms of sin( u ). Specifically, cos(2 u ) = 1 − 2sin²( u ).

How does that help? Well, 4 A is 2×2 A , isn’t it?

cos(2 u ) = 1 − 2sin²( u )

cos(2×2 A ) = 1 − 2sin²(2 A )

cos(4 A ) = 1 − 2sin²(2 A )

That transforms the original equation to

1 − 2sin²(2 A ) − sin(2 A ) = 0

which can be rewritten in standard form as

2sin²(2 A ) + sin(2 A ) − 1 = 0

Now you have the equation in terms of only one function (sine) and only one angle (2 A ).

Now that the equation involves only a single function of a single angle, your next task is to solve for that function value.

Example A continues. Recapping what was done so far,

cos(4 A ) − sin(2 A ) = 0  ⇒

You want to solve for sin(2 A ). You should recognize that the equation is really a quadratic,

2 y ² + y − 1 = 0, where y = sin(2 A )

It can be factored in a straightforward way, as ( y +1)(2 y −1), which gives

(sin(2 A ) + 1) (2sin(2 A ) − 1) = 0

From algebra you know that if a product is 0 then you solve by setting each factor to 0:

sin(2 A ) + 1 = 0       or       2sin(2 A ) − 1 = 0

sin(2 A ) = −1       or       sin(2 A ) = 1/2

Example B :

3tan²( B /2) − 1 = 0

tan²( B /2) = 1/3

and then take square root of both sides:

tan( B /2) = ±√ 1/3  = ±√ 3 /3

It’s important to remember to use the plus-or-minus sign ± when taking the square root of both sides; otherwise you could overlook some solutions.

After solving for a function value, now you solve for the angle. If it’s a multiple of π/6 (30°) or π/4 (45°), you can easily solve it exactly. If it’s a half-multiple of those angles, you may be able to use the half-angle formulas to write an exact solution. Otherwise, you’ll need to write the solution as an arcfunction.

Trig equations have one important difference from other types of equations. Trig functions are periodic , meaning that they repeat their values over and over. Therefore a trig equation has an infinite number of solutions if it has any.

Think about an equation like sin  u  = 1. π/2 is a solution, but the sine function repeats its values every 2π. Therefore π/2±2π, π/2±4π, and so on are equally good solutions. To show this, write the solution as u  = π/2 + 2π n , where n is understood to be any integer, positive, negative, or zero. (The tangent and cotangent functions repeat all their values every π radians, so the solution to tan  v  = 1 is v  = π/4 + π n , not +2π n .)

2sin²(2 A ) + sin(2 A ) − 1 = 0 ⇒

The sine of 3π/2 is −1, so the first possibility reduces to 2 A  = 3π/2. But remember that the sine function is periodic, so write

sin(2 A ) = −1 ⇒ 2 A  = 3π/2 + 2π n .

For the second possibility, sin(2 A ) = 1/2, there are two solutions, because sin(π/6) and sin(5π/6) both equal 1/2, and again we add 2π n to the angle to account for all solutions:

sin(2 A ) = 1/2 ⇒ 2 A = π/6 + 2π n or 5π/6 + 2π n

Combining these, here are the three solutions for the original equation:

2 A = 3π/2 + 2π n    or    π/6 + 2π n     or    5π/6 + 2π n

Example B continues. Recapping what was done so far,

3tan²( B /2) − 1 = 0 ⇒

tan( B /2) = ±√ 3 /3

What angle has a tangent value of √ 3 /3? the angle π/6. And where does the tangent have a value of −√ 3 /3? at the angle 5π/6. This gives the solutions

B /2 = π/6 + π n    or     5π/6 + π n

Remember that the tangent and cotangent have period π and not 2π.

Example C :

Of course, you don’t always luck out with nice angles. Take a look at this equation:

sec(3 C ) = 2.5

1/sec(3 C ) = 1/2.5

cos(3 C ) = 0.4

For what angles is that true? We write arccos(0.4) to mean the angle in quadrant I that has a cosine equal to 0.4. (Some books write cos −1 (0.4) instead of arccos(0.4). I prefer the arccos notation because the superscript −1 makes many students think of 1/cos(0.4), which has a different meaning entirely.)

So initially we would write 3 C  = arccos(0.4) + 2π n . But that’s not the whole story: any angle in quadrant I has a reflection in quadrant IV with the same cosine value, so we need to account for both angles:

3 C = arccos(0.4) + 2π n   or   2π−arccos(0.4) + 2π n

Note that the base angle is always nonnegative and less than 2π: 2π−arccos(0.4) + 2π n , not simply −arccos(0.4) + 2π n . This is necessary to make step 5 come out right.

Now it’s time to abandon angular thinking and go for the variable. As you will see, it is very important to do this step after step 3 , not before. 2π n or π n must be added to the angle, not the variable, to reflect the period of the trig function.

sin(2 A ) = −1   or   sin(2 A ) = 1/2 ⇒

2 A = 3π/2 + 2π n   or   π/6 + 2π n    or   5π/6 + 2π n

Now divide both sides by 2 to solve for the variable, A :

A = 3π/4 + π n   or   π/12 + π n   or   5π/12 + π n

Be sure to divide the entire equation, so that the 2π n becomes π n .

Why is the order of steps so important? The 2πn came in because the sine function has a period of 2π: if you take an angle and add 2π to it, it looks like the same angle and all six of its function values are unchanged. But now we’re no longer dealing with the angle 2 A , we’re dealing with the variable A . In this equation, we say that adding π to any solution for A will give another solution for A .

For instance, set n =1 and obtain solutions A  = 7π/4, 13π/12, or 17π/12. True, sin(7π/4) doesn’t equal −1. But the equation was sin(2 A ) = −1, not sin( A ) = −1. If you substitute A  = 7π/4 in sin(2 A ), you get sin(2·7π/4) = sin(7π/2) which does equal −1. Always pay attention to whether you’re dealing with the angle or the variable.

tan( B /2) = ±√ 3 /3 ⇒

B /2 = π/6 + π n   or   5π/6 + π n

Multiplying both sides by 2 gives

B = π/3 + 2π n   or   5π/3 + 2π n

Once again, the angle was B /2 and had a period of π; the variable is B and has a period of 2π.

Example C continues. Recapping what was done so far,

sec(3 C ) = 2.5 ⇒

Divide both sides by 3:

C = (1/3)arccos(0.4) + (2π/3) n   or   (2π/3)−(1/3)arccos(0.4) + (2π/3) n

It’s a matter of taste whether to combine terms in that second solution:

C = (1/3)arccos(0.4) + (2π/3) n    or    −(1/3)arccos(0.4) + (2 n +1)π/3

Does the problem specify a solution interval for the variable? Sometimes this is done in interval notation, like [0;2π); other times it’s done as an inequality, 0 <=  x  < 2π. If no restriction is given, you should give all real solutions as shown in step 4 .

But if solutions are restricted to a particular interval, you have a bit more work to do after solving for the variable .

2 A = 3π/2 + 2π n   or   π/6 + 2π n    or   5π/6 + 2π n ⇒

Now suppose that only solutions on the interval [0;2π) were wanted.

The general solutions have a period of π (from +π n ), and therefore there will be two cycles between 0 and 2π:

The solutions for n  = 2 are all larger than the 2π boundary of the interval. There are six solutions to the equation for A within the interval [0;2π). In order, they are π/12, 5π/12, 3π/4, 13π/12, 17π/12, and 7π/4.

B /2 = π/6 + π n   or   5π/6 + π n ⇒

Suppose that the problem specified solutions between 0 and π/2. As you can see, even with n  = 0 there is just one solution within the limits [0;π/2). Answer: π/3 only.

3 C = arccos(0.4) + 2π n   or   2π−arccos(0.4) + 2π n ⇒

Suppose again that limits of C in [0;2π) are stated. The period of the variable is 2π/3, so there are three cycles ( n =0,1,2) between 0 and 2π. Here are the values, with decimal approximations:

Since 2π is about 6.28, there are six solutions within the stated limits. C  = about 0.3864, 0.6608, 2.4808, 2.7552, 4.5752, and 4.8496.

Example D :

Solve ½sin(2 D ) − ½ = 0 for D in [0;2π).

Solution: There is only one function (sin) of one angle (2 D ), so step 1 is complete. Don’t “simplify” ½sin(2 D ) to sin( D ): that is not a valid operation . And don’t convert sin(2 D ) to 2 sin( D ) cos( D ); while that is mathematically valid, it makes the equation more complicated, not simpler.

Step 2 : solve for the function.

½sin(2 D ) = ½

sin(2 D ) = 1

Step 3 : find the angle.

2 D = π/2 + 2π n

Step 4 : solve for the variable.

D = π/4 + π n

Step 5 : apply any restrictions. The period of the variable is π, which means there will be two cycles in the interval [0;2π).

Example E :

Find all solutions for sin²( E /2) − cos²( E /2) = 1.

Solution: In step 1 , you use identities to get the equation in term of one function of one angle. As often happens in trig, you have a choice of which identity you want to use. You could replace cos²( E /2) with 1−sin²( E /2), or you could use the double-angle formula cos(2 u ) = cos²( u ) − sin²( u ) with u  =  E /2. That second approach leads to a simpler equation:

sin²( E /2) − cos²( E /2) = 1

cos²( E /2) − sin²( E /2) = −1

cos(2 E /2) = −1

cos( E ) = −1

Step 3 : write down the angle:

E = π + 2π n = (2 n +1)π

Step 4 is already done, since the angle is the variable.

Step 5 is easily done: the problem specifically asks for all values of E .

Answer: E  =(2 n +1)π for any integer n .

Showed an intermediate step in a “straightforward” solution.

Add the possibility of angles being multiples of half of a special angle .

Converted page from HTML 4.01 to HTML5, and aitalicized the variable names.

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• 20 Apr 2002 : First publication.

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## Learning Objectives

• Verify the fundamental trigonometric identities.
• Simplify trigonometric expressions using algebra and the identities.

In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.

In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

## Verifying the Fundamental Trigonometric Identities

Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.

To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities , the even-odd identities, the reciprocal identities, and the quotient identities.

We will begin with the Pythagorean identities (Table $$\PageIndex{1}$$), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.

The second and third identities can be obtained by manipulating the first. The identity $$1+{\cot}^2 \theta={\csc}^2 \theta$$ is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: $$1+{\cot}^2 \theta={\csc}^2 \theta$$

\begin{align*} 1+{\cot}^2 \theta&= (1+\dfrac{{\cos}^2}{{\sin}^2})\qquad \text{Rewrite the left side}\\ &= \left(\dfrac{{\sin}^2}{{\sin}^2}\right)+\left (\dfrac{{\cos}^2}{{\sin}^2}\right)\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\sin}^2+{\cos}^2}{{\sin}^2}\\ &= \dfrac{1}{{\sin}^2}\\ &= {\csc}^2 \end{align*}

Similarly,$$1+{\tan}^2 \theta={\sec}^2 \theta$$can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives

\begin{align*} 1+{\tan}^2 \theta&= 1+{\left(\dfrac{\sin \theta}{\cos \theta}\right )}^2\qquad \text{Rewrite left side}\\ &= {\left (\dfrac{\cos \theta}{\cos \theta}\right )}^2+{\left (\dfrac{\sin \theta}{\cos \theta}\right)}^2\qquad \text{Write both terms with the common denominator}\\ &= \dfrac{{\cos}^2 \theta+{\sin}^2 \theta}{{\cos}^2 \theta}\\ &= \dfrac{1}{{\cos}^2 \theta}\\ &= {\sec}^2 \theta \end{align*}

Recall that we determined which trigonometric functions are odd and which are even. The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle (Table $$\PageIndex{2}$$).

Recall that an odd function is one in which $$f(−x)= −f(x)$$ for all $$x$$ in the domain off. f. The sine function is an odd function because $$\sin(−\theta)=−\sin \theta$$. The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of $$\dfrac{\pi}{2}$$ and $$−\dfrac{\pi}{2}$$. The output of $$\sin\left (\dfrac{\pi}{2}\right )$$ is opposite the output of $$\sin \left (−\dfrac{\pi}{2}\right )$$. Thus,

\begin{align*} \sin\left (\dfrac{\pi}{2}\right)&=1 \\[4pt] \sin\left (-\dfrac{\pi}{2}\right) &=-\sin\left (\dfrac{\pi}{2}\right) \\[4pt] &=-1 \end{align*}

This is shown in Figure $$\PageIndex{2}$$.

Recall that an even function is one in which

$$f(−x)=f(x)$$ for all $$x$$ in the domain of $$f$$

The graph of an even function is symmetric about the y- axis. The cosine function is an even function because $$\cos(−\theta)=\cos \theta$$. For example, consider corresponding inputs $$\dfrac{\pi}{4}$$ and $$−\dfrac{\pi}{4}$$. The output of $$\cos\left (\dfrac{\pi}{4}\right)$$ is the same as the output of $$\cos\left (−\dfrac{\pi}{4}\right)$$. Thus,

\begin{align*} \cos\left (−\dfrac{\pi}{4}\right ) &=\cos\left (\dfrac{\pi}{4}\right) \\[4pt] &≈0.707 \end{align*}

See Figure $$\PageIndex{3}$$.

For all $$\theta$$ in the domain of the sine and cosine functions, respectively, we can state the following:

• Since $$\sin(−\theta)=−\sin \theta$$,sine is an odd function.
• Since $$\cos(−\theta)=\cos \theta$$,cosine is an even function.

The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity,$$\tan(−\theta)=−\tan \theta$$. We can interpret the tangent of a negative angle as

$\tan (−\theta)=\dfrac{\sin (−\theta)}{\cos (−\theta)}=\dfrac{−\sin \theta}{\cos \theta}=−\tan \theta. \nonumber$

Tangent is therefore an odd function, which means that $$\tan(−\theta)=−\tan(\theta)$$ for all $$\theta$$ in the domain of the tangent function.

The cotangent identity, $$\cot(−\theta)=−\cot \theta$$,also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as

$\cot(−\theta)=\dfrac{\cos(−\theta)}{\sin(−\theta)}=\dfrac{\cos \theta}{−\sin \theta}=−\cot \theta.\nonumber$

Cotangent is therefore an odd function, which means that $$\cot(−\theta)=−\cot(\theta)$$ for all $$\theta$$ in the domain of the cotangent function.

The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as

$\csc(−\theta)=\dfrac{1}{\sin(−\theta)}=\dfrac{1}{−\sin \theta}=−\csc \theta. \nonumber$

The cosecant function is therefore odd.

Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as

$\sec(−\theta)=\dfrac{1}{\cos(−\theta)}=\dfrac{1}{\cos \theta}=\sec \theta. \nonumber$

The secant function is therefore even.

To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.

The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. (Table $$\PageIndex{3}$$). Recall that we first encountered these identities when defining trigonometric functions from right angles in Right Angle Trigonometry .

The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities (Table $$\PageIndex{4}$$).

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

## SUMMARIZING TRIGONOMETRIC IDENTITIES

The Pythagorean identities are based on the properties of a right triangle.

${\cos}^2 \theta+{\sin}^2 \theta=1$

$1+{\cot}^2 \theta={\csc}^2 \theta$

$1+{\tan}^2 \theta={\sec}^2 \theta$

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.

$\tan(−\theta)=−\tan \theta$

$\cot(−\theta)=−\cot \theta$

$\sin(−\theta)=−\sin \theta$

$\csc(−\theta)=−\csc \theta$

$\cos(−\theta)=\cos \theta$

$\sec(−\theta)=\sec \theta$

The reciprocal identities define reciprocals of the trigonometric functions.

$\sin \theta=\dfrac{1}{\csc \theta}$

$\cos \theta=\dfrac{1}{\sec \theta}$

$\tan \theta=\dfrac{1}{\cot \theta}$

$\csc \theta=\dfrac{1}{\sin \theta}$

$\sec \theta=\dfrac{1}{\cos \theta}$

$\cot \theta=\dfrac{1}{\tan \theta}$

The quotient identities define the relationship among the trigonometric functions.

$\tan \theta=\dfrac{\sin \theta}{\cos \theta}$

$\cot \theta=\dfrac{\cos \theta}{\sin \theta}$

## Example $$\PageIndex{1}$$: Graphing the Equations of an Identity

Graph both sides of the identity $$\cot \theta=\dfrac{1}{\tan \theta}$$. In other words, on the graphing calculator, graph $$y=\cot \theta$$ and $$y=\dfrac{1}{\tan \theta}$$.

See Figure $$\PageIndex{4}$$.

We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.

## How to: Given a trigonometric identity, verify that it is true.

• Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.
• Look for opportunities to factor expressions, square a binomial, or add fractions.
• Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
• If these steps do not yield the desired result, try converting all terms to sines and cosines.

## Example $$\PageIndex{2}$$: Verifying a Trigonometric Identity

Verify $$\tan \theta \cos \theta=\sin \theta$$.

We will start on the left side, as it is the more complicated side:

\begin{align*} \tan \theta \cos \theta &=\left(\dfrac{\sin \theta}{\cos \theta}\right)\cos \theta \\[4pt] &=\sin \theta \end{align*}

This identity was fairly simple to verify, as it only required writing $$\tan \theta$$ in terms of $$\sin \theta$$ and $$\cos \theta$$.

## Exercise $$\PageIndex{1}$$

Verify the identity $$\csc \theta \cos \theta \tan \theta=1$$.

\begin{align*} \csc \theta \cos \theta \tan \theta=\left(\dfrac{1}{\sin \theta}\right)\cos \theta\left(\dfrac{\sin \theta}{\cos \theta}\right) \\[4pt] & =\dfrac{\cos \theta}{\sin \theta}(\dfrac{\sin \theta}{\cos \theta}) \\[4pt] & =\dfrac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[4pt] &=1 \end{align*}

## Example $$\PageIndex{3A}$$: Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

$$(1+\sin x)[1+\sin(−x)]={\cos}^2 x$$

Working on the left side of the equation, we have

$$(1+\sin x)[1+\sin(−x)]=(1+\sin x)(1-\sin x)$$

\begin{align*} \sin(-x)&= -\sin x \\ [5pt] &=1-{\sin}^2 x\qquad \text{Difference of squares} \\ [5pt] &={\cos}^2 x \\ {\cos}^2 x&= 1-{\sin}^2 x \\ \end{align*}

## Example $$\PageIndex{3B}$$: Verifying a Trigonometric Identity Involving $${\sec}^2 \theta$$

Verify the identity $$\dfrac{{\sec}^2 \theta−1}{{\sec}^2 \theta}={\sin}^2 \theta$$

As the left side is more complicated, let’s begin there.

\begin{align*} \dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{({\tan}^2 \theta +1)-1}{{\sec}^2 \theta}\\ {\sec}^2 \theta&= {\tan}^2 \theta +1\\ &= \dfrac{{\tan}^2 \theta}{{\sec}^2 \theta}\\ &= {\tan}^2 \theta\left (\dfrac{1}{{\sec}^2 \theta}\right )\\ &= {\tan}^2 \theta \left ({\cos}^2 \theta\right )\\ {\cos}^2 \theta&= \dfrac{1}{{\sec}^2 \theta}\\ &= \left (\dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\right )\\ {\tan}^2 \theta&= \dfrac{{\sin}^2 \theta}{{\cos}^2 \theta}\\ &= {\sin}^2 \theta \end{align*}

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

\begin{align*} \dfrac{{\sec}^2 \theta-1}{{\sec}^2 \theta}&= \dfrac{{\sec}^2 \theta}{{\sec}^2 \theta}-\dfrac{1}{{\sec}^2 \theta}\\ &= 1-{\cos}^2 \theta\\ &= {\sin}^2 \theta \end{align*}

In the first method, we used the identity $${\sec}^2 \theta={\tan}^2 \theta+1$$ and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.

## Exercise $$\PageIndex{2}$$

Show that $$\dfrac{\cot \theta}{\csc \theta}=\cos \theta$$.

\begin{align*} \dfrac{\cot \theta}{\csc \theta}&= \dfrac{\tfrac{\cos \theta}{\sin \theta}}{\dfrac{1}{\sin \theta}}\\ &= \dfrac{\cos \theta}{\sin \theta}\cdot \dfrac{\sin \theta}{1}\\ &= \cos \theta \end{align*}

## Example $$\PageIndex{4}$$: Creating and Verifying an Identity

Create an identity for the expression $$2 \tan \theta \sec \theta$$ by rewriting strictly in terms of sine.

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

\begin{align*} 2 \tan \theta \sec \theta&= 2\left (\dfrac{\sin \theta}{\cos \theta}\right )\left(\dfrac{1}{\cos \theta}\right )\\ &= \dfrac{2\sin \theta}{{\cos}^2 \theta}\\ &= \dfrac{2\sin \theta}{1-{\sin}^2 \theta}\qquad \text{Substitute } 1-{\sin}^2 \theta \text{ for } {\cos}^2 \theta \end{align*}

$$2 \tan \theta \sec \theta=\dfrac{2 \sin \theta}{1−{\sin}^2 \theta}$$

## Example $$\PageIndex{5}$$: Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

$$\dfrac{{\sin}^2(−\theta)−{\cos}^2(−\theta)}{\sin(−\theta)−\cos(−\theta)}=\cos \theta−\sin \theta$$

\begin{align*} \dfrac{{\sin}^2(-\theta)-{\cos}^2(-\theta)}{\sin(-\theta)-\cos(-\theta)}&= \dfrac{{[\sin(-\theta)]}^2-{[\cos(-\theta)]}^2}{\sin(-\theta)-\cos(-\theta)}\\ &= \dfrac{{(-\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta} \;\; \; , \sin(-x) = -\sin\space x\text { and } \cos(-x)=\cos \space x\\ &= \dfrac{{(\sin \theta)}^2-{(\cos \theta)}^2}{-\sin \theta -\cos \theta}\qquad \text{Difference of squares}\\ &= \dfrac{(\sin \theta-\cos \theta)(\sin \theta+\cos \theta)}{-(\sin \theta+\cos \theta)}\\ &= \cos \theta-\sin \theta \end{align*}

## Exercise $$\PageIndex{3}$$

Verify the identity $$\dfrac{{\sin}^2 \theta−1}{\tan \theta \sin \theta−\tan \theta}=\dfrac{\sin \theta+1}{\tan \theta}$$.

\begin{align*} \dfrac{{\sin}^2 \theta-1}{\tan \theta \sin \theta-\tan \theta}&= \dfrac{(\sin \theta +1)(\sin \theta -1)}{\tan \theta(\sin \theta -1)}\\ &= \dfrac{\sin \theta+1}{\tan \theta} \end{align*}

## Example $$\PageIndex{6}$$: Verifying an Identity Involving Cosines and Cotangents

Verify the identity: $$(1−{\cos}^2 x)(1+{\cot}^2 x)=1$$.

\begin{align*} (1-{\cos}^2 x)(1+{\cot}^2 x)&= (1-{\cos}^2 x)\left(1+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right)\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x}{{\sin}^2 x}+\dfrac{{\cos}^2 x}{{\sin}^2 x}\right )\qquad \text{Find the common denominator}\\ &= (1-{\cos}^2 x)\left(\dfrac{{\sin}^2 x +{\cos}^2 x}{{\sin}^2 x}\right)\\ &= ({\sin}^2 x)\left (\dfrac{1}{{\sin}^2 x}\right )\\ &= 1 \end{align*}

## Using Algebra to Simplify Trigonometric Expressions

We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.

For example, the equation $$(\sin x+1)(\sin x−1)=0$$ resembles the equation $$(x+1)(x−1)=0$$, which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.

Another example is the difference of squares formula, $$a^2−b^2=(a−b)(a+b)$$, which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.

## Example $$\PageIndex{7A}$$: Writing the Trigonometric Expression as an Algebraic Expression

Write the following trigonometric expression as an algebraic expression: $$2{\cos}^2 \theta+\cos \theta−1$$.

Notice that the pattern displayed has the same form as a standard quadratic expression,$$ax^2+bx+c$$. Letting $$\cos \theta=x$$,we can rewrite the expression as follows:

$$2x^2+x−1$$

This expression can be factored as $$(2x+1)(x−1)$$. If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for $$x$$. At this point, we would replace $$x$$ with $$\cos \theta$$ and solve for $$\theta$$.

## Example $$\PageIndex{7B}$$: Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression using the difference of squares: $$4{cos}^2 \theta−1$$.

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares.

\begin{align*} 4{\cos}^2 \theta-1&= {(2\cos \theta)}^2-1\\ &= (2\cos \theta-1)(2\cos \theta+1) \end{align*}

If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let $$\cos \theta=x$$, rewrite the expression as $$4x^2−1$$, and factor $$(2x−1)(2x+1)$$. Then replace $$x$$ with $$\cos \theta$$ and solve for the angle.

## Exercise $$\PageIndex{4}$$

Rewrite the trigonometric expression using the difference of squares: $$25−9{\sin}^2 \theta$$.

This is a difference of squares formula: $$25−9{\sin}^2 \theta=(5−3\sin \theta)(5+3\sin \theta)$$.

## Example $$\PageIndex{8}$$: Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:

$${\csc}^2 \theta−{\cot}^2 \theta$$

\begin{align*} 1+{\cot}^2 \theta&= {\csc}^2 \theta\\ \text{Now we can simplify by substituting } 1+{\cot}^2 \theta \text{ for } {\csc}^2 \theta\\ {\csc}^2 \theta-{\cot}^2 \theta&= 1+{\cot}^2 \theta-{\cot}^2 \theta\\ &= 1 \end{align*}

## Exercise $$\PageIndex{5}$$

Use algebraic techniques to verify the identity: $$\dfrac{\cos \theta}{1+\sin \theta}=\dfrac{1−\sin \theta}{\cos \theta}$$.

(Hint: Multiply the numerator and denominator on the left side by $$1−\sin \theta$$.)

\begin{align*} \dfrac{\cos \theta}{1+\sin \theta}\left(\dfrac{1-\sin \theta}{1-\sin \theta}\right)&= \dfrac{\cos \theta (1-\sin \theta)}{1-{\sin}^2 \theta}\\ &= \dfrac{\cos \theta (1-\sin \theta)}{{\cos}^2 \theta}\\ &= \dfrac{1-\sin \theta}{\cos \theta} \end{align*}

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

• Fundamental Trigonometric Identities
• Verifying Trigonometric Identities

## Key Equations

Key concepts.

• There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
• Graphing both sides of an identity will verify it. See Example $$\PageIndex{1}$$.
• Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example $$\PageIndex{2}$$ and Example $$\PageIndex{3}$$.
• The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See Example $$\PageIndex{4}$$.
• We can create an identity and then verify it. See Example $$\PageIndex{5}$$.
• Verifying an identity may involve algebra with the fundamental identities. See Example $$\PageIndex{6}$$ and Example $$\PageIndex{7}$$.
• Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See Example $$\PageIndex{8}$$, Example $$\PageIndex{9}$$, and Example $$\PageIndex{10}$$.

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## Unit 4: Trigonometric equations and identities

• Intro to arcsine (Opens a modal)
• Intro to arctangent (Opens a modal)
• Intro to arccosine (Opens a modal)
• Restricting domains of functions to make them invertible (Opens a modal)
• Domain & range of inverse tangent function (Opens a modal)
• Using inverse trig functions with a calculator (Opens a modal)
• Inverse trigonometric functions review (Opens a modal)
• Trigonometric equations and identities: FAQ (Opens a modal)
• Evaluate inverse trig functions Get 3 of 4 questions to level up!

## Sinusoidal equations

• Solving sinusoidal equations of the form sin(x)=d (Opens a modal)
• Cosine equation algebraic solution set (Opens a modal)
• Cosine equation solution set in an interval (Opens a modal)
• Sine equation algebraic solution set (Opens a modal)
• Solving cos(θ)=1 and cos(θ)=-1 (Opens a modal)
• Solve sinusoidal equations (basic) Get 3 of 4 questions to level up!
• Solve sinusoidal equations Get 3 of 4 questions to level up!

## Sinusoidal models

• Interpreting solutions of trigonometric equations (Opens a modal)
• Trig word problem: solving for temperature (Opens a modal)
• Trigonometric equations review (Opens a modal)
• Interpret solutions of trigonometric equations in context Get 3 of 4 questions to level up!
• Sinusoidal models word problems Get 3 of 4 questions to level up!

• Trig angle addition identities (Opens a modal)
• Using the cosine angle addition identity (Opens a modal)
• Using the cosine double-angle identity (Opens a modal)
• Proof of the sine angle addition identity (Opens a modal)
• Proof of the cosine angle addition identity (Opens a modal)
• Proof of the tangent angle sum and difference identities (Opens a modal)
• Using the trig angle addition identities Get 3 of 4 questions to level up!

## Using trigonometric identities

• Finding trig values using angle addition identities (Opens a modal)
• Using the tangent angle addition identity (Opens a modal)
• Using trig angle addition identities: finding side lengths (Opens a modal)
• Using trig angle addition identities: manipulating expressions (Opens a modal)
• Using trigonometric identities (Opens a modal)
• Trig identity reference (Opens a modal)
• Find trig values using angle addition identities Get 3 of 4 questions to level up!

## Challenging trigonometry problems

• Trig challenge problem: area of a triangle (Opens a modal)
• Trig challenge problem: area of a hexagon (Opens a modal)
• Trig challenge problem: cosine of angle-sum (Opens a modal)
• Trig challenge problem: arithmetic progression (Opens a modal)
• Trig challenge problem: maximum value (Opens a modal)
• Trig challenge problem: multiple constraints (Opens a modal)
• Trig challenge problem: system of equations (Opens a modal)

Calcworkshop

## Solving Trig Equations with 3 Simple Steps!

// Last Updated: January 21, 2020 - Watch Video //

We are now going to combine our Algebra skills with our knowledge of Trig Identities and the Unit Circle and begin Solving Trigonometric Equations !

What makes these equations different than plain-old equations, is that they are conditional.

What’s a conditional equation?

An equation that is true for some, but not all values.

Don’t worry, this may sound tricky, but in all honesty, it’s very straightforward as long as you remember your Left Hand Trick (i.e., Reference Angles) from our amazing Unit Circle!

The main objective, as in solving any equation, is to isolate our trig function, using known algebraic methods such as addition, subtraction, factoring, etc., and then ask ourselves one very important question…

“What angle on the unit circle yields this length or coordinate?”

That’s it!

Truth be told, there are only two to three steps for Solving any Trigonometric Equation, and we are going to walk ourselves through this process with countless examples, just like the one you see below.

Solve the trig equation for all values of x

Also, we are going to look at an incredibly powerful trick for solving multiple angle or half-angle equations – it takes the sting right out of it, I promise!

Yes, as Purple Math states, you will need to put your thinking cap on, because sometimes your first instinct doesn’t get you any closer to a desired outcome, but taking a different approach will – don’t give up.

Moreover, as SOS Math points out, sometimes we will encounter solutions that are extraneous, or invalid, and that is why we must always check our answers.

The overall process is direct and uncomplicated, and with some practice, will become straightforward and clear.

## Solving Trigonometric Equations – Video

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## Solving Simple (to Medium-Hard) Trig Equations

Easy/Medium Hard

Solving trig equations use both the reference angles and trigonometric identities that you've memorized, together with a lot of the algebra you've learned. Be prepared to need to think in order to solve these equations.

In what follows, it is assumed that you have a good grasp of the trig-ratio values in the first quadrant , how the unit circle works, the relationship between radians and degrees , and what the various trig functions' curves look like, at least on the first period. If you're not sure of yourself, go back and review those topics first.

Content Continues Below

## Solve sin( x ) + 2 = 3 over the interval 0° ≤  x  < 360°

Just as with linear equations , I'll first isolate the variable-containing term:

sin( x ) + 2 = 3

sin( x ) = 1

Now I'll use the reference angles I've memorized to get my final answer.

So, in degrees, my answer is:

x = 90°

## Solve tan 2 (θ) + 3 = 0 on the interval 0° ≤ θ < 360°

There's the temptation to quickly recall that the tangent of 60° involves the square root of 3 and slap down an answer, but this equation doesn't actually have a solution. I can see this when I slow down and do the steps. My first step is:

tan 2 (θ) = −3

Can any square (of a tangent, or of any other trig function) be negative ? No! So my answer is:

no solution

## Solve katex.render("\\boldsymbol{\\color{green}{\\small{ 2 \\cos^2(x) - \\sqrt{3\\,} \\cos(x) = 0 }}}", typed01); on the interval 0° ≤  x  < 360°

The left-hand side of this equation factors. I'm used to doing simple factoring like this:

2 y 2 + 3 y = 0

y (2 y + 3) = 0

...and then solving each of the factors. The same sort of thing works here. To solve the equation they've given me, I will start with the factoring:

I've done the algebra; that is, I've done the factoring and then I've solved each of the two factor-related equations. This created two trig equations. So now I can do the trig; namely, solving those two resulting trigonometric equations, using what I've memorized about the cosine wave. From the first equation, I get:

cos( x ) = 0:

x = 90°, 270°

From the second equation, I get:

2 cos( x ) = sqrt[3]:

x = 30°, 330°

Putting these the two solution sets together, I get the solution for the original equation as being:

x = 30°, 90°, 270°, 330°

## Solve sin 2 (θ) − sin(θ) = 2 on the interval 0 ≤ θ < 2π

First, I'll get everything over to one side of the "equals" sign:

sin 2 (θ) − sin(θ) − 2 = 0

This equation is "a quadratic in sine"; that is, the form of the equation is the quadratic-equation format:

a X 2 + b X + c = 0

In the case of the equation they're wanting me to solve, X = sin(θ) , a  = 1 , b  = −1 , and c  = −2 .

Since this is quadratic in form, I can apply some quadratic-equation methods. In the case of this equation, I can factor the quadratic:

( sin(θ) − 2 )( sin(θ) + 1 ) = 0

The first factor gives me the related trig equation:

sin(θ) = 2

But the sine is never more than 1 , so this equation is not solvable; it has no solution.

The other factor gives me the second related trig equation:

sin(θ) + 1 = 0

sin(θ) = −1

θ = (3/2)π

(If you're doing degrees-only solutions in your class, the solution value above equates to "270°".)

## Solve cos 2 (α) + cos(α) = sin 2 (α) on the interval 0° ≤ x < 360°

I can use a trig identity to get a quadratic in cosine:

cos 2 (α) + cos(α) = sin 2 (α)

cos 2 (α) + cos(α) = 1 − cos 2 (α)

2cos 2 (α) + cos(α) − 1 = 0

( 2cos(α) − 1 )( cos(α) + 1 ) = 0

cos(α) = 1/2, cos(α) = −1

α = 60°, 180°, 300°

## Solve sin(β) = sin(2β) on the interval 0° ≤ β < 360°

I can use a double-angle identity on the right-hand side, and rearrange and simplify; then I'll factor:

sin(β) = 2sin(β)cos(β)

sin(β) − 2sin(β) cos(β) = 0

sin(β) ( 1 − 2cos(β) ) = 0

sin(β) = 0,cos(β) = 1/2

The sine wave (from the first trig equation) is zero at 0° , 180° , and 360° . But, in the original exercise, 360° is not included, so this last solution value doesn't count, in this particular instance.

β = 0°, 60°, 180°, 300°

## Solve sin( x ) + cos( x ) = 1 on the interval 0° ≤ x < 360°

Hmm... I'm really not seeing anything here. It sure would have been nice if one of these trig expressions were squared...

Well, why don't I square both sides, and see what happens?

( sin( x ) + cos( x ) ) 2 = (1) 2

sin 2 ( x ) + 2sin( x )cos( x ) + cos 2 ( x ) = 1

[ sin 2 ( x + cos 2 ( x ) ] + 2sin( x )cos( x ) = 1

1 + 2sin( x )cos( x ) = 1

2sin( x )cos( x ) = 0

sin( x )cos( x ) = 0

Huh; go figure: I squared, and got something that I could work with. Nice!

From the last line above, either sine is zero or else cosine is zero, so my solution appears to be:

x = 0°, 90°, 180°, 270°

However (and this is important!), I squared to get this solution, and squaring is an "irreversible" process.

(Why? If you square something, you can't just square-root to get back to what you'd started with, because the squaring may have changed a sign somewhere.)

So, to be sure of my results, I need to check my answers in the original equation, to make sure that I didn't accidentally create solutions that don't actually count. Plugging back in, I see:

sin(0°) + cos(0°) = 0 + 1 = 1

...so the solution " x = 0° " works

sin(90°) + cos(90°) = 1 + 0 = 1

...so the solution " x = 90° " works, too

sin(180°) + cos(180°) = 0 + (−1) = −1

...oh, okay, so " x = 180° " does NOT work

sin(270°) + cos(270°) = (−1) + 0 = −1

...so " x = 270° " doesn't work, either

It's a good thing that I checked my solutions, because two of them don't actually work. They were created by the process of squaring.

My actual solution is:

x = 0°, 90°

Note: In the above, I could have stopped at this line:

...and used the double-angle identity for sine, in reverse, instead of dividing off the 2 in the next-to-last line in my computations. The answer would have been the same, but I would have needed to account for the solution interval:

2sin( x )cos( x ) = sin(2 x ) = 0

Then 2 x = 0°, 180°, 360°, 540° , etc, and dividing off the 2 from the x would give me x = 0°, 90°, 180°, 270° , which is the same almost-solution as before. After doing the necessary check (because of the squaring) and discarding the extraneous solutions, my final answer would have been the same as previously.

The squaring trick in the last example above doesn't come up often, but if nothing else is working, it might be worth a try. Keep it in mind for the next test.

URL: https://www.purplemath.com/modules/solvtrig.htm

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## How to Solve Trigonometric Equations?

A trigonometric equation is an equation whose variable is expressed in terms of the value of the trigonometric function. In this step-by-step guide, you will learn more about trigonometric equations and solving them.

## A step-by-step guide to solving trigonometric equations

Trigonometric equations are similar to algebraic equations and can be linear equations, quadratic equations, or polynomial equations.

To solve trigonometric equations, we use some general results and solutions of trigonometric equations. These results are as follows:

• For any real numbers $$x$$ and $$y$$, $$sin\: x = sin\: y$$ implies $$x=n\pi +\left(-1\right)^ny$$, where $$n ∈ Z$$.
• For any real numbers $$x$$ and $$y$$, $$cos\: x = cos\: y$$ implies $$x = 2nπ ± y$$, where $$n ∈ Z$$.
• If $$x$$ and $$y$$ are not odd multiples of $$\frac{π}{2}$$, then $$tan\: x = tan\: y$$ implies $$x = nπ + y$$, where $$n ∈ Z$$.

## Solving trigonometric equations

We have two types of solutions to the trigonometric equations:

• Principal Solution:

The initial values of angles for trigonometric functions are called principal solutions.

• General Solution:

The values of the angles for the same answer of the trigonometric function are called the general solution of the trigonometric function. The general solutions of $$sin\:θ$$, $$cos\:θ$$, and $$tan\:θ$$ are as follows.

• $$sin\:θ = sin\:α$$, and the general solution is $$θ=n\pi +\left(-1\right)^nα$$, where $$n ∈ Z$$
• $$cos\:θ = cos\:α$$, and the general solution is $$θ = 2nπ + α$$, where $$n ∈ Z$$
• $$tan\:θ = tan\:α$$, and the general solution is $$θ = nπ + α$$, where $$n ∈ Z$$

## Steps to solve trigonometric equations

To solve a trigonometric equation, the following steps should be followed.

• Convert the given trigonometric equation to an equation with a single trigonometric ratio (sin, cos, tan).
• Change the equation with the trigonometric equation, having multiple angles, or submultiple angles into a simple angle.
• Now express the equation as a polynomial equation, a quadratic equation, or a linear equation.
• Solve the trigonometric equation similar to normal equations and find the value of the trigonometric ratio.
• The angle of the trigonometric ratio or the value of the trigonometric ratio shows the solution of the trigonometric equation.

## Solving Trigonometric Equations – Example 1:

Find the principal solutions of the equation $$tan x=-\frac{1}{\sqrt{3}}$$.

We know that $$tan\left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$$

$$tan\:\left(\pi \:-\:\frac{\pi }{6}\:\right)=-tan\:\left(\frac{\pi }{6}\right)=-\:\frac{1}{\sqrt{3}}$$

$$tan\:\left(2\pi \:-\:\frac{\pi }{6}\right)=-tan\left(\frac{\pi }{6}\right)=-\:\frac{1}{\sqrt{3}}$$

So, the principal solutions are $$tan\: (π – \frac{π}{6}) = tan\: (\frac{5π}{6})$$ and $$tan\: (2π – \frac{π}{6}) = tan\: (\frac{11π}{6})$$.

## Exercises for Solving Trigonometric Equations

Solve each trigonometric function for all possible values in degree..

• $$\color{blue}{cos\:x+\sqrt{3}=-cos\:x}$$
• $$\color{blue}{sin\:2x-\frac{\sqrt{3}}{2}=0}$$
• $$\color{blue}{4\:sin\:\theta -1=2\:sin\:\theta +1}$$
• $$\color{blue}{csc\:x+cot\:x=1}$$
• $$\color{blue}{\:x=150^{\circ \:}+360^{\circ \:}n,\:x=210^{\circ \:}+360^{\circ \:}n}$$
• $$\color{blue}{x=30^{\circ \:}+180^{\circ \:}n,\:x=60^{\circ \:}+180^{\circ \:}n}$$
• $$\color{blue}{\:θ=90^{\circ \:}+360^{\circ \:}n}$$
• $$\color{blue}{\:x=90^{\circ \:}+360^{\circ \:}n}$$

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## Trigonometric Functions

3.8 solving trigonometric equations, learning objectives.

In this section, you will:

• Solve linear trigonometric equations in sine and cosine.
• Solve equations involving a single trigonometric function.
• Solve trigonometric equations using a calculator.
• Solve trigonometric equations that are quadratic in form.
• Solve trigonometric equations using fundamental identities.
• Solve trigonometric equations with multiple angles.

Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill)

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

## Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all.  Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid.

Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions.  The period of both the sine function and the cosine function is $2\pi.$ In other words, every $2\pi$ units, the y- values repeat, so  $\mathrm{sin}\left(\theta\right)=\mathrm{sin}\left(\theta \pm2k\pi\right)$.  If we need to find all possible solutions, then we must add $2\pi k,$ where $k$ is an integer, to the initial solution.

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

## Example 1:  Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation $\mathrm{cos}\left(\theta\right) =\frac{1}{2}.$

From the unit circle , we know that there will be two angles where $\mathrm{cos}\left(\theta\right) =\frac{1}{2}$ in one complete revolution, i.e. $0\le\theta\le{2\pi}.$ They will occur in the first and fourth quadrants. We recognize $\frac{1}{2}$ as a value from one of our special right triangles.  We can identify the acute angle as $\frac{\pi}{3}.$  We can then use this as the reference angle to find the angle in the fourth quadrant by computing $2\pi-\frac{\pi}{3}.$

These are the solutions in the interval $\left[0,2\pi \right].$ All possible solutions are given by  $\frac{\pi }{3}\pm2k\pi$ and $\frac{5\pi }{3}\pm2k\pi$ where $k$ is an integer.

## Example 2:  Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation $\mathrm{sin}\left(t\right)=\frac{1}{2}.$

Solving for all possible values of t means that solutions include angles beyond the period of $2\pi .$ From previous work with the unit circle, we know that there are two solutions in one revolution.  Since the value is positive, we know these solutions are in the first and second quadrants.  From our special right triangles, we know that the angle in the first quadrant is $\frac{\pi }{6}$ and using this as the reference angle, the solution in the second quadrant is $\frac{5\pi }{6}.$  But the problem is asking for all possible values that solve the equation.

Therefore, the answer is  $\frac{\pi }{6}\pm2\pi k$ and $\frac{5\pi }{6}\pm2\pi k$  where $k$ is an integer.

## Example 3:  Solve the Trigonometric Equation in Linear Form

Solve the equation exactly: $2\text{ }\mathrm{cos}\left(\theta\right) -3=-5,\text{ }\text{ }0\le \theta <2\pi .$

Use algebraic techniques to solve the equation.

Solve exactly the following linear equation on the interval $\left[0,2\pi \right):\text{ }2\text{ }\mathrm{sin}\left(x\right)+1=0.$

$x=\frac{7\pi }{6},\text{ }\frac{11\pi }{6}$

## Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and information we know from the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the equation in terms of the reciprocal function, and solve for the angles using the functions we are most familiar with. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is $\pi,$ not $2\pi.$ Further, the domain of tangent is all real numbers with the exception of odd integer multiples of $\frac{\pi }{2},$ unless, of course, a problem places its own restrictions on the domain.

## Example 4:  Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: $\mathrm{csc}\left(\theta\right) =-2,\text{ }0\le \theta <4\pi .$

We want all values of $\theta$ for which $\mathrm{csc}\left(\theta\right) =-2$ over the interval $0\le \theta <4\pi.$

## Example 5:  Solving an Equation Involving Tangent

Solve the equation exactly: $\mathrm{tan}\left(\theta -\frac{\pi }{2}\right)=1,\text{ } 0\le \theta <2\pi .$

Recall that the tangent function has a period of $\pi.$ On the interval $\left[0,\pi \right),$ and at the angle of $\frac{\pi }{4},$ the tangent has a value of 1. However, the angle we want is $\left(\theta -\frac{\pi }{2}\right).$ Thus, if $\mathrm{tan}\left(\frac{\pi }{4}\right)=1,$ then

Over the interval $\left[0,2\pi \right),$ we have two solutions:

Find all solutions for $\mathrm{tan}\left(x\right)=\sqrt[\leftroot{1}\uproot{2} ]{3}.$

$\frac{\pi }{3}\pm\pi k$

## Example 6:  Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation $2\left(\mathrm{tan}\left(x\right)+3\right)=5+\mathrm{tan}\left(x\right),\text{ }0\le x<2\pi .$

We can solve this equation using only algebra. Isolate the expression $\mathrm{tan}\left(x\right)$ on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of $-1:\theta =\frac{3\pi }{4}$ and $\theta =\frac{7\pi }{4}.$

## Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

## Example 7:  Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation $\mathrm{sin}\left(\theta\right) =0.8,$ where $\theta$ is in radians.

Make sure mode is set to radians. To find $\theta,$ use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the $\mathrm{sin}^{-1}$ function. What is shown on the screen is $\mathrm{sin}^{-1}\left(\text{ }\right).$ The calculator is ready for the input within the parentheses. For this problem, we enter $\mathrm{sin}^{-1}\left(0.8\right),$ and press ENTER. Thus, to four decimals places,

The solution is $0.9273\pm2\pi k.$

Keep in mind that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine.  Since the sine value is positive, there will be another solution in quadrant 2.  The other angle is obtained by using $\pi -\theta.$

This gives us a second set of values in the form $2.2143\pm2\pi k.$

The angle measurements in degrees are based on the first revolution angles of

$\theta\approx {53.1}^{\circ}\text{ or } \theta\approx {180}^{\circ}-{53.1}^{\circ}\approx {126.9}^{\circ}$

## Example 8:  Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation $\mathrm{sec}\left(\theta\right) =-4,$ giving your answer in radians.

We can begin with some algebra.

Check that the MODE is in radians. Now use the inverse cosine function.

Since $\frac{\pi }{2}\approx 1.57$ and $\pi \approx 3.14,$ we know that 1.8235 is between these two numbers, thus $\theta \approx \text{1}\text{.8235}$ is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 1 .

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is $\theta^{\prime} \approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}$ The other solution in quadrant III is $\pi +\text{1}\text{.3181}\approx \text{4}\text{.4597.}$

The solutions are $1.8235\pm2\pi k$ and $4.4597\pm2\pi k.$

Solve $\mathrm{csc}\left(\theta\right) =3.$

$\theta \approx 0.33984\pm2\pi k$ and $\theta \approx 2.80176\pm2\pi k$

## Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as $x$ or $u.$ If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Given a trigonometric equation, solve using algebra .

• Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
• Substitute the trigonometric expression with a single variable, such as $x$ or $u.$
• Solve the equation the same way an algebraic equation would be solved.
• Substitute the trigonometric expression back in for the variable in the resulting expressions.
• Solve for the angle.

## Example 9:  Solving a Trigonometric Equation in Quadratic Form Using the Square Root Property

Solve the problem exactly: $2\text{ }{\mathrm{sin}}^{2}\left(\theta\right) -1=0, \text{ }0\le \theta <2\pi .$

## Example 10: Solving a Trigonometric Equation in Quadratic Form Using the Quadratic Equation

Solve the equation exactly: $\mathrm{cos}^{2}\left(\theta\right)+3\mathrm{cos}\left(\theta\right)-1=0,\text{ }0\le \theta <2\pi .$

We begin by using substitution and replacing $\mathrm{cos}\left(\theta\right)$ with $u.$ It is not necessary to use substitution, but it may make the problem easier to solve visually. Let $\mathrm{cos}\left(\theta\right) =u.$ We have

The equation cannot be factored, so we will use the quadratic formula $u=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a},$ where $a=1,\text{ }b=3$ and $c=-1.$

Replace $u$ with $\mathrm{cos}\left(\theta\right),$ and solve. Thus,

Note that only the + sign is used. This is because we get an error when we solve $\theta =\mathrm{cos}^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)$ on a calculator, since the domain of the inverse cosine function is $\left[-1,1\right].$ Therefore the solution is

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

## Example 11:  Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: $2\text{ }\mathrm{sin}^{2}\left(\theta\right) -5\text{ }\mathrm{sin}\left(\theta\right)+3=0,\text{ }0\le \theta \le 2\pi .$

Now set each factor equal to zero and solve the two equations $2\text{ }\mathrm{sin}\left(\theta\right) -3=0$ and $\mathrm{sin}\left(\theta\right) -1=0.$

For the first equation,

The first equation did not have any solution.  To solve the second equation,

The only solution for this equation is $\theta=\frac{\pi}{2}.$

Make sure to check all solutions on the given domain as some factors have no solution. This is because the range of the sine function is $\left[-1,1\right].$

## Example 12: Solving an Equation Using an Identity

Solve the equation exactly using an identity: $3\text{ }\mathrm{cos}\left(\theta\right)+3=2\text{ }{\mathrm{sin}}^{2}\left(\theta\right) ,\text{ }0\le \theta <2\pi .$

If we rewrite the right side, we can write the equation in terms of cosine.  Recall that the Pythagorean Identity is $\mathrm{sin}^2\left(\theta\right)+\mathrm{cos}^2\left(\theta\right)=1$ and can be solved for $\mathrm{sin}^{2}\left(\theta\right)$ so $\mathrm{sin}^{2}\left(\theta\right)=1-\mathrm{cos}^{2}\left(\theta\right).$

Again, remember that there are two quadrants where the $\mathrm{cos}\left(\theta\right)=-\frac{1}{2}.$  These are quadrants 2 and 3.  We can find the acute reference angle of $\frac{\pi}{3}$ and use that to generate the solutions in these quadrants.

Our solutions are $\frac{2\pi}{3},\text{ }\frac{4\pi }{3},\text{ }\pi.$

Solve $\mathrm{sin}^{2}\left(\theta\right)=2\mathrm{cos}\left(\theta\right)+2,\text{ }0\le\theta \le2\pi$ [Hint: Make a substitution to express the equation only in terms of cosine.]

$\mathrm{cos}\left(\theta\right) =-1,\text{ }\theta =\pi$

Solve the quadratic equation $2\mathrm{cos}^{2}\left(\theta\right) +\mathrm{cos}\left(\theta\right) =0.$

$\frac{\pi }{2},\text{ }\frac{2\pi }{3},\text{ }\frac{4\pi }{3},\text{ }\frac{3\pi }{2}$

## Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

We will need to use some new identities in this section.  They are called the  double-angle identities. We will not take the time to show where the following identities come from. Keep in mind that there are other trigonometric identities that we have not covered in this material. If you are interested, you can look up the sum and difference formulas for sine and cosine, and use those to generate some other identities, including the ones shown below.

The double angle identities are:

$\mathrm{sin}\left(2\theta\right)=2\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)$

$\\$

\begin{align*}\mathrm{cos}\left(2\theta\right)&=\mathrm{cos}^{2}\left(\theta\right)-\mathrm{sin}^{2}\left(\theta\right)\text{ or,}\\&=1-2\mathrm{sin}^{2}\left(\theta\right)\text{ or,}\\&=2\mathrm{cos}^{2}\left(\theta\right)-1\end{align*}

## Example 13: Solving the Equation Using a Double-Angle Formula

Solve the equation exactly using a double-angle formula: $\mathrm{cos}\left(2\theta\right)=\mathrm{cos}\left(\theta\right).$

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

Set the factors equal to zero and solve.

\begin{align*}2\mathrm{cos}\left(\theta\right)+1&=0&\mathrm{cos}\left(\theta\right)-1&=0\\\mathrm{cos}\left(\theta\right)&=-\frac{1}{2}&\mathrm{cos}\left(\theta\right)&=1\\\theta&=\frac{2\pi}{3}\pm2\pi k&\theta&=0\pm2\pi k\\\theta&=\frac{4\pi}{3}\pm2\pi k\end{align*}

## Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as $\mathrm{sin}\left(2x\right)$ or $\mathrm{cos}\left(3x\right).$ When confronted with these equations, recall that $y=\mathrm{sin}\left(2x\right)$ is a horizontal compression by a factor of 2 of the function $y=\mathrm{sin}\left(x\right).$  On an interval of $2\pi,$ we can graph two periods of $y=\mathrm{sin}\left(2x\right),$ as opposed to one cycle of $y=\mathrm{sin}\left(x\right).$ This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to $\mathrm{sin}\left(2x\right)=0$ compared to $\mathrm{sin}\left(x\right)=0.$ This information will help us solve the similar type of equation shown in the example.

## Example 14: Solving a Multiple Angle Trigonometric Equation

Solve exactly: $\mathrm{cos}\left(2x\right)=\frac{1}{2}$ on $\left[0,2\pi \right).$

We can see that this equation is the standard equation with a multiple of an angle. If $\mathrm{cos}\left(\theta\right)=\frac{1}{2},$ we know $\theta$ is in quadrants I and IV. While $\theta =\mathrm{cos}^{-1}\left(\frac{1}{2}\right)$ will only yield solutions in quadrants I and II because of the range of the inverse cosine function, we recognize that the solutions to the equation $\mathrm{cos}\left(\theta\right)=\frac{1}{2}$ will be in quadrants I and IV using ideas from our unit circle.

Therefore, the possible angles are $\theta=\frac{\pi }{3}$ and $\theta=\frac{5\pi}{3}.$ So, $2x=\frac{\pi}{3}$ or $2x=\frac{5\pi}{3},$ which means that $x=\frac{\pi}{6}$ or $x=\frac{5\pi}{6}.$ $\\$Does this make sense? Yes, because $\mathrm{cos}\left(2\left(\frac{\pi}{6}\right)\right)=\mathrm{cos}\left(\frac{\pi}{3}\right)=\frac{1}{2}.$

In quadrant I, $2x=\frac{\pi}{3},$ so $x=\frac{\pi}{6}$ as noted. Let us revolve around the circle again:

One more rotation yields

$x=\frac{13\pi}{6}>2\pi,$ so this value for $x$ is larger than $2\pi,$ so it is not a solution on $\left[0,2\pi \right).$

In quadrant IV, $2x=\frac{5\pi }{3},$ so $x=\frac{5\pi }{6}$ as noted. Let us revolve around the circle again:

$x=\frac{17\pi }{6}>2\pi ,$ so this value for $x$ is larger than $2\pi,$ which means it is not a solution on $\left[0,2\pi \right).$

Our solutions are $\frac{\pi}{6},\text{ }\frac{5\pi }{6},\text{ }\frac{7\pi }{6},$ and $\frac{11\pi }{6}.$ Note that whenever we solve a problem in the form of $\mathrm{sin}\left(nx\right)=c$ or $\mathrm{cos}\left(nx\right)=c,$ we must go around the unit circle ${n}$ times.

We can see this easily if we first write the solution for $2x$ in the general form.  One of these equations is shown below:

\begin{align*}2x&=\frac{\pi}{3}+2\pi\cdot k\\x&=\frac{\pi}{6}+\pi \cdot k&&\text{Divide both sides by 2.}\end{align*}$\\$

We can now see that adding $\pi$ to $\frac{\pi}{6}$ gives us $\frac{7\pi}{6}$ and that adding $2\cdot\pi$ would give us $\frac{13\pi}{6}$  which is larger than $2\pi.$

Access these online resources for additional instruction and practice with solving trigonometric equations.

• Solving Trigonometric Equations I
• Solving Trigonometric Equations II
• Solving Trigonometric Equations III
• Solving Trigonometric Equations IV
• Solving Trigonometric Equations V
• Solving Trigonometric Equations VI

## Key Concepts

• When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.
• Equations involving a single trigonometric function can be solved or verified using the unit circle.
• We can also solve trigonometric equations using a graphing calculator.
• Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.
• We can also use the identities to solve trigonometric equation.
• We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval.
• Solving Trigonometric Equations. Authored by : Douglas Hoffman. Provided by : Openstax. Located at : https://cnx.org/contents/[email protected]:aeVxcRIM@7/Solving-Trigonometric-Equations . Project : Essential Precalcus, Part 2. License : CC BY: Attribution

• Mastering Trigonometric Equations: A Step-by-Step Guide

Welcome to Warren Institute! In this article, we will dive into the world of trigonometric equations and learn how to solve them step by step. Trigonometric equations involve trigonometric functions such as sine, cosine, and tangent, and solving them can sometimes be a daunting task. But fear not! We will break it down for you, providing clear explanations and easy-to-follow examples. By the end of this article, you will have a solid understanding of the process of solving trigonometric equations and be able to tackle them confidently. Let's get started on our journey to conquer trigonometry !

## Introduction to Trigonometric Equations

Steps to solve trigonometric equations, common mistakes to avoid, applications of trigonometric equations, how can i solve trigonometric equations using the unit circle, what are the different methods for solving trigonometric equations algebraically, can you explain the steps involved in solving a trigonometric equation graphically, are there any specific strategies or tips for solving trigonometric equations involving multiple angles, can you provide a step-by-step example of solving a complex trigonometric equation.

In this section, we will provide an overview of trigonometric equations and their importance in mathematics education . We will explain the basic concepts and properties related to solving these equations.

This section will outline the step-by-step process for solving trigonometric equations. We will discuss various techniques such as factoring, using trigonometric identities, and applying inverse trigonometric functions. Each step will be clearly explained and demonstrated with examples.

In this section, we will highlight some common errors that students often make while solving trigonometric equations. We will identify these mistakes and provide tips on how to avoid them. Understanding these pitfalls will help students become more proficient in solving trigonometric equations.

Trigonometric equations have numerous real-life applications in fields such as physics, engineering, and architecture. In this section, we will explore some practical examples where solving trigonometric equations is essential. By understanding these applications, students will develop a deeper appreciation for the relevance of trigonometry in various professional disciplines.

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## Trigonometric Equations Calculator

Get detailed solutions to your math problems with our trigonometric equations step-by-step calculator . practice your math skills and learn step by step with our math solver. check out all of our online calculators here .,  example,  solved problems,  difficult problems.

Solved example of trigonometric equations

The reciprocal sine function is cosecant: $\frac{1}{\csc(x)}=\sin(x)$

Move everything to the left hand side of the equation

Combining like terms $8\sin\left(x\right)$ and $-4\sin\left(x\right)$

Factor the polynomial $4\sin\left(x\right)-2$ by it's greatest common factor (GCF): $2$

Divide both sides of the equation by $2$

Simplifying the quotients

Divide $0$ by $2$

Intermediate steps

We need to isolate the dependent variable $x$, we can do that by simultaneously subtracting $-1$ from both sides of the equation

$x+0=x$, where $x$ is any expression

Multiply $-1$ times $-1$

Divide $1$ by $2$

The angles where the function $\sin\left(x\right)$ is $\frac{1}{2}$ are

The angles expressed in radians in the same order are equal to

Struggling with math.

Access detailed step by step solutions to thousands of problems, growing every day!

## What are the steps to solve trigonometric equations?

The steps to solve the trigonometric equations are: reduce the given trigonometric equation in terms of the basic trigonometric ratios. factorize the given trigonometric polynomial in terms of the ratio. solve each factor and write down the general solution. example: tan 2 x + 1 = sin 2 x cos 2 x + 1 ∵ sin x cos x = tan x ⇒ tan 2 x + 1 = sin 2 x cos 2 x + cos 2 x cos 2 x ⇒ tan 2 x + 1 = sin 2 x + cos 2 x cos 2 x ⇒ tan 2 x + 1 = 1 cos 2 x ∵ sin 2 x + cos 2 x = 1 ⇒ tan 2 x + 1 = sec 2 x ∵ 1 cos x = sec x.

#### IMAGES

1. How to Solve Trigonometric Equations: A Simple Tutorial

3. Solving Trigonometric Equations Involving a Squared Function Involving

4. how to solve trigonometric equations step by step

5. How to Solve Trigonometric Equations: A Simple Tutorial

6. How To Solve Trigonometric Equations With Multiple Angles

#### VIDEO

1. Trigonometric Equations

2. Simple Trigonometric Equation

3. Trigonometric equations

4. Solving Trigonometric Equations

5. How to: Solve Trigonometric Equations

6. How to solve trigonometric equation using different trigonometric identity

1. How to Solve Trigonometric Equations: A Simple Tutorial

1 Know the Solving concept. [1] To solve a trig equation, transform it into one or many basic trig equations. Solving trig equations finally results in solving 4 types of basic trig equations. 2 Know how to solve basic trig equations. [2] There are 4 types of basic trig equations: sin x = a ; cos x = a tan x = a ; cot x = a

2. Trigonometric Equations

Some of the examples of trigonometric equations are as follows. Sin2x - Sin4x + Sin6x = 0 2Cos 2 x + 3Sinx = 0 Cos4x = Cos2x Sin2x + Cosx = 0 Sec 2 2x = 1 - Tan2x Trigonometric Equations Formulas We use some results and general solutions of the basic trigonometric equations to solve other trigonometric equations. These results are as follows:

3. Solving Trigonometric Equations By Finding All Solutions

4.9K 537K views 6 years ago New Trigonometry Playlist This trigonometry video provides a basic introduction into solving trigonometric equations. it explains how to find all solutions by...

4. Solving Trigonometric Equations 1

Solving Trigonometric Equations 1 corbettmaths 186K subscribers Subscribe Subscribed Like 102K views 4 years ago AQA Level 2 Further Maths This video shows how to solve trig equations...

5. 3.3: Solving Trigonometric Equations

We begin by factoring: 2x2 + x = 0 x(2x + 1) = 0 Set each factor equal to zero. x = 0 2x + 1 = 0 x = − 1 2. Then, substitute back into the equation the original expression sinθ for x. Thus, sinθ = 0 θ = 0, π sinθ = − 1 2 θ = 7π 6, 11π 6. The solutions within the domain 0 ≤ θ < 2π are θ = 0, π, 7π 6, 11π 6.

6. How to Solve Trigonometric Equations

Overview Step 1. Get one function of one angle. Step 2. Solve for the value (s) of a trig function. Step 3. Solve for the angle. Step 4. Solve for the variable. Step 5. Apply any restrictions. More Examples What's New Overview If a trig equation can be solved analytically, these steps will do it:

7. 9.1: Solving Trigonometric Equations with Identities

The Pythagorean identities are based on the properties of a right triangle. cos2θ + sin2θ = 1. 1 + cot2θ = csc2θ. 1 + tan2θ = sec2θ. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle. tan( − θ) = − tanθ. cot( − θ) = − cotθ.

8. Calculus I

Example 1 Solve 2cos(t) =√3 2 cos ( t) = 3 . Show Solution Now, in a calculus class this is not a typical trig equation that we'll be asked to solve. A more typical example is the next one. Example 2 Solve 2cos(t) =√3 2 cos ( t) = 3 on [−2π,2π] [ − 2 π, 2 π] . Show Solution So, let's see if you've got all this down.

9. Trigonometric equations and identities

In this unit, you'll explore the power and beauty of trigonometric equations and identities, which allow you to express and relate different aspects of triangles, circles, and waves. You'll learn how to use trigonometric functions, their inverses, and various identities to solve and check equations and inequalities, and to model and analyze problems involving periodic motion, sound, light, and ...

10. How to Solve Trig Equations (with 3 Simple Steps!)

The main objective, as in solving any equation, is to isolate our trig function, using known algebraic methods such as addition, subtraction, factoring, etc., and then ask ourselves one very important question… "What angle on the unit circle yields this length or coordinate?" That's it!

11. Solving trig equations is not as bad as it looks!

Purplemath. Solving trig equations use both the reference angles and trigonometric identities that you've memorized, together with a lot of the algebra you've learned. Be prepared to need to think in order to solve these equations.. In what follows, it is assumed that you have a good grasp of the trig-ratio values in the first quadrant, how the unit circle works, the relationship between ...

12. How to Solve Trigonometric Equations?

General Solution: The values of the angles for the same answer of the trigonometric function are called the general solution of the trigonometric function. The general solutions of \ (sin\:θ\), \ (cos\:θ\), and \ (tan\:θ\) are as follows. \ (sin\:θ = sin\:α\), and the general solution is \ (θ=n\pi +\left (-1\right)^nα\), where \ (n ∈ Z\)

13. 3.8 Solving Trigonometric Equations

In other words, trigonometric equations may have an infinite number of solutions. The period of both the sine function and the cosine function is 2π. In other words, every 2π units, the y- values repeat, so sin(θ) = sin(θ ± 2kπ) . If we need to find all possible solutions, then we must add 2πk, where k is an integer, to the initial solution.

14. Trigonometric Equation Calculator

How to solve trigonometric equations step-by-step? To solve a trigonometric simplify the equation using trigonometric identities. Then, write the equation in a standard form, and isolate the variable using algebraic manipulation to solve for the variable. Use inverse trigonometric functions to find the solutions, and check for extraneous solutions.

15. Mastering Trigonometric Equations: A Step-by-Step Guide

Here's a step-by-step example of solving a complex trigonometric equation: Step 1: Identify the given equation and determine the range of values for which you want to solve it. Step 2: Simplify the equation by applying trigonometric identities or algebraic manipulations if necessary.

16. How to Solve Trigonometric Equations (Precalculus

A very In-Depth look into solving equations that involve trig functions. We will focus on solving equations without having to use inverse trigonometric funct...

17. Basic trigonometric equations

Solving trigonometric equations Basic trigonometric equations. Trigonometric equations can be solved in degrees or radians using CAST and its period to find other solutions within the range ...

18. PDF Solving Trigonometric Equations

Solving Trigonometric Equations. 2cos 1 0. x. − = 2cos 1. x = cos 1 2. x = Solve: 5, 33. x. π π = Step 1: Isosolate cos . x . using algebraic skills. Step 2: Determine in which quadrants cosine is positive. Use the inverse. function to assist by finding the angle in Quad I first. Then use that angle. as the reference angle for the other ...

19. Step-by-Step Calculator

Step-by-Step Calculator Trigonometry Full pad Examples Frequently Asked Questions (FAQ) Is there a step by step calculator for math? Symbolab is the best step by step calculator for a wide range of math problems, from basic arithmetic to advanced calculus and linear algebra.

20. Trigonometric Equations Calculator & Solver

Trigonometric Equations Calculator Get detailed solutions to your math problems with our Trigonometric Equations step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here. 8sin ( x) = 2 + 4 csc ( x) Go! Math mode Text mode . ( ) / ÷ 2 √ √ ∞ e π ln log

21. What are the steps to solve trigonometric equations?

The steps to solve the trigonometric equations are: Reduce the given trigonometric equation in terms of the basic trigonometric ratios. Factorize the given trigonometric polynomial in terms of the ratio. Solve each factor and write down the general solution. Example: tan 2 x + 1 = sin 2 x cos 2 x + 1 ∵ sin x cos x = tan x

22. Solving Trig Equations: Using the Unit Circle

http://www.mathpowerline.com Solving trig equations is made a lot easier if you know how to use the unit circle. These problems are when the angle me...

23. Solving Trigonometric Equations STEP BY STEP.

Step by step how to solve trig functiions easily