AP®MathsCollege BoardPrecalculusStudy GuidesTrigonometric & Polar FunctionsTrigonometric Equations, Inequalities & IdentitiesSolving Equations with Trigonometric Identities

Solving Equations with Trigonometric Identities (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

Solving equations with trigonometric identities

When are trigonometric identities useful for solving equations?

Some trigonometric equations cannot be solved directly
- because they involve multiple different trigonometric functions
- or arguments of different sizes (e.g. both $\sin x$ and $\sin (2 x)$ )
In these cases, applying a trigonometric identity can rewrite the equation in a more accessible form
- typically one that involves only a single trig function or only a single angle measure
Once the equation has been rewritten, it can be solved using the standard techniques for simple trigonometric equations
The key idea is that
- an equivalent analytic representation
- can make the structure of the equation easier to work with

What are common situations where identities help?

Equations mixing single-angle and double-angle terms
- e.g. $\sin (2 x) - \cos x = 0$
- Apply the double-angle identity $\sin (2 x) = 2 \sin x \cos x$ to convert to a single-angle equation
  - then factor
Equations mixing sine-squared and cosine-squared (or sine and cosine-squared, or cosine and sine squared)
- e.g. $2 \cos^{2} x + \sin x - 1 = 0$
- Apply the Pythagorean identity to express everything in terms of one function (here, sine)
  - then treat it as a quadratic
Equations involving reciprocal trig functions
- e.g. $\sec^{2} x = 3 \tan x - 1$
- Apply the rearranged Pythagorean identity ( $\sec^{2} x = 1 + \tan^{2} x$ ) to express everything in terms of tangent

What is the general process?

Start by identifying the obstacle that prevents direct solution
- Usually this will be multiple trig functions or multiple angle measures appearing in the same equation
Then choose an identity that, when applied, will eliminate the obstacle
- If multiple angles appear, use sum/difference/double-angle identities to bring everything to a single angle
- If multiple functions appear, use the Pythagorean identity (or a quotient/reciprocal relationship) to bring everything to a single function
This allows you to apply the identity and rewrite the equation
- and then solve the resulting equation using the standard techniques
  - i.e. factoring, finding principal solutions, using symmetry/periodicity
Finally verify that all solutions lie within the specified solution interval
- and that no additional solutions still need to be found

Worked Example

Solve the equation $\sin (2 x) - \cos x = 0$ for values of $x$ in the interval $[0, 2 π)$ .

Answer

The equation involves both $\sin (2 x)$ and $\cos x$

i.e. different angles in the trig functions

Apply the double-angle identity $\sin (2 x) = 2 \sin x \cos x$ to convert to a single angle

$2 \sin x \cos x - \cos x = 0$

Factor out the common factor $\cos x$

$\cos x (2 \sin x - 1) = 0$

This gives two simpler equations

$\cos x = 0 or 2 \sin x - 1 = 0$

First solve $\cos x = 0$ on $[0, 2 π)$

From the unit circle, cosine is zero when the terminal ray is vertical:

$x = \frac{π}{2}, x = \frac{3 π}{2}$

Then solve $2 \sin x - 1 = 0$ on $[0, 2 π)$

Adding 1 to both sides then dividing by 2 gives

$\sin x = \frac{1}{2}$

Find the principal solution

$x = \sin^{- 1} (\frac{1}{2}) = \frac{π}{6}$

Use the symmetry of the sine function to find the other solution in the interval

or $x = π - \frac{π}{6} = \frac{5 π}{6}$

Combine all solutions for your final answer

$x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$

Worked Example

Solve the equation $2 \cos^{2} x + \sin x - 1 = 0$ for values of $x$ in the interval $[0, 2 π)$ .

Answer

The equation involves both $\cos^{2} x$ and $\sin x$

Apply the Pythagorean identity in the form $\cos^{2} x = 1 - \sin^{2} x$ to convert everything to sine:

$2 (1 - \sin^{2} x) + \sin x - 1 = 0$

Expand and simplify

$2 - 2 \sin^{2} x + \sin x - 1 = 0$

$- 2 \sin^{2} x + \sin x + 1 = 0$

Multiply through by $- 1$ to make the leading coefficient positive

This isn't absolutely necessary, but it makes the algebra in the next steps simpler

$2 \sin^{2} x - \sin x - 1 = 0$

That is a 'hidden quadratic' in $\sin x$ , which can be factored

If you find it tricky to do this with the equation in terms of $\sin x$ , you could also use the substitution $y = \sin x$ , and do the factoring in terms of $y$ before converting back

$(2 \sin x + 1) (\sin x - 1) = 0$

This gives two simpler equations

$2 \sin x + 1 = 0 or \sin x - 1 = 0$

First solve $2 \sin x + 1 = 0$ on $[0, 2 π)$

$2 \sin x + 1 = 0 \Rightarrow \sin x = - \frac{1}{2}$

Find the principal value

$x = \sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$

That value is not in $[0, 2 π)$ , so add $2 π$ to it to find a value that is

$x = - \frac{π}{6} + 2 π = \frac{11 π}{6}$

Use the symmetry of the sine function to find the other value in the solution interval
- $\frac{11 π}{6}$ is $\frac{π}{6}$ less than $2 π$ , so the other value will be $\frac{π}{6}$ more than $π$

$x = π + \frac{π}{6} = \frac{7 π}{6}$

So the two solutions in $[0, 2 π)$ are

$x = \frac{7 π}{6}, \frac{11 π}{6}$

Now solve $\sin x - 1 = 0$ on $[0, 2 π)$ :

$\sin x - 1 = 0 \Rightarrow \sin x = 1$

From the unit circle, sine is 1 only when the terminal ray is vertical and extending up from the origin

$x = \frac{π}{2}$

Combine all solutions for your final answer

$x = \frac{π}{2}, \frac{7 π}{6}, \frac{11 π}{6}$

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

Solving Equations with Trigonometric Identities (College Board AP® Precalculus): Study Guide

Solving equations with trigonometric identities

When are trigonometric identities useful for solving equations?

What are common situations where identities help?

What is the general process?

Worked Example

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis