Solving Trig Equations with Identities

Solving trig equations with identities means rewriting an equation so it has only one trig function of one angle, then solving it. For $\sin 2x = \cos x$ , replace $\sin 2x$ with $2 \sin x \cos x$ , factor, and set each factor equal to 0. Use Pythagorean, double-angle and reciprocal identities to make the swap.

Before this lesson: You should know Solving Trig Equations, Fundamental Trig Identities and Double-Angle Formulas.

Why Use Identities to Solve Trig Equations?

An equation like $2 \cos^2 x + 3 \sin x - 3 = 0$ mixes sine and cosine. You can’t solve for one function while the other is still there. An identity lets you trade one for the other.

Here, $\cos^2 x = 1 - \sin^2 x$ turns the whole equation into sine only. Then it is a quadratic in $\sin x$ , and you already know how to solve those.

The same idea works for different angles. In $\sin 2x = \cos x$ , the angles $2x$ and $x$ don’t match. The double-angle formula rewrites $\sin 2x$ with angle $x$ , so both sides use the same angle.

Graph with x from 0 to 2 pi and y from −1.5 to 1.5. The solid blue curve y = sin 2x and the dashed orange curve y = cos x cross at four points, marked with green dots, at x = pi/6, pi/2, 5 pi/6 and 3 pi/2. — The graphs of y = sin 2x and y = cos x cross four times on [0, 2π), so sin 2x = cos x has four solutions.

How to Solve Trig Equations with Identities

Look for mixed functions (like sine and cosine) or mixed angles (like $2x$ and $x$ ).
Pick an identity that leaves one function of one angle. Common choices are in the table below.
Move every term to one side so the other side is 0.
Factor, then set each factor equal to 0.
Solve each simpler equation on $[0, 2\pi)$ .
If you squared both sides, check every answer in the original equation.

If the equation has…	Try this identity	Result
$\sin^2 x$ and $\cos x$	$\sin^2 x = 1 - \cos^2 x$	Only cosine
$\cos^2 x$ and $\sin x$	$\cos^2 x = 1 - \sin^2 x$	Only sine
$\sin 2x$ with $\sin x$ or $\cos x$	$\sin 2x = 2 \sin x \cos x$	Only angle $x$
$\cos 2x$ and $\cos x$	$\cos 2x = 2\cos^2 x - 1$	Only cosine
$\cos 2x$ and $\sin x$	$\cos 2x = 1 - 2\sin^2 x$	Only sine
$\sec^2 x$ and $\tan x$	$\sec^2 x = 1 + \tan^2 x$	Only tangent

Worked Examples

Example 1: Use a Pythagorean identity

Solve $2 \cos^2 x + 3 \sin x - 3 = 0$ on $[0, 2\pi)$ .

2(1 - \sin^2 x) + 3 \sin x - 3 = 0

Replace $\cos^2 x$ with $1 - \sin^2 x$ (see Pythagorean identities)

2 - 2\sin^2 x + 3 \sin x - 3 = 0

Distribute the 2.

-2\sin^2 x + 3 \sin x - 1 = 0

Combine the constants, $2 - 3 = -1$ .

2\sin^2 x - 3 \sin x + 1 = 0

Multiply both sides by −1.

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

Factor the trinomial (see factoring trinomials when a is not 1)

2 \sin x - 1 = 0

Set the first factor equal to 0.

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

Add 1 and divide by 2.

Solve it on the unit circle: $x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$

\sin x - 1 = 0

Set the second factor equal to 0.

\sin x = 1

Add 1 to both sides.

x = \frac{\pi}{2}

Solve it on the unit circle.

Answer: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}$

Example 2: Use the double-angle formula for sine

Solve $\sin 2x = \cos x$ on $[0, 2\pi)$ .

2 \sin x \cos x = \cos x

Replace $\sin 2x$ with $2 \sin x \cos x$ .

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

Subtract $\cos x$ from both sides.

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

Factor out the GCF $\cos x$ (see factoring out the GCF)

Set the first factor equal to 0: $\cos x = 0$ , so $x = \frac{\pi}{2}$ or $x = \frac{3\pi}{2}$

2 \sin x - 1 = 0

Set the second factor equal to 0.

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

Add 1 and divide by 2.

Solve it: $x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$

Answer: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ . These are the four crossing points in the graph above. Don’t divide both sides by $\cos x$ ; that would lose $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

Example 3: Use the double-angle formula for cosine

Solve $\cos 2x + 3 \cos x + 2 = 0$ on $[0, 2\pi)$ .

2\cos^2 x - 1 + 3 \cos x + 2 = 0

The other terms use $\cos x$ , so replace $\cos 2x$ with $2\cos^2 x - 1$ .

2\cos^2 x + 3 \cos x + 1 = 0

Combine the constants, $-1 + 2 = 1$ .

(2 \cos x + 1)(\cos x + 1) = 0

Factor the trinomial.

2 \cos x + 1 = 0

Set the first factor equal to 0.

\cos x = -\frac{1}{2}

Subtract 1 and divide by 2.

Cosine is negative in Quadrants II and III: $x = \frac{2\pi}{3}$ or $x = \frac{4\pi}{3}$

\cos x + 1 = 0

Set the second factor equal to 0.

\cos x = -1

Subtract 1 from both sides.

x = \pi

Solve it on the unit circle.

Answer: $x = \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$

Example 4: Square both sides, then check

Solve $\sin x + \cos x = 1$ on $[0, 2\pi)$ .

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

Square both sides.

\sin^2 x + 2 \sin x \cos x + \cos^2 x = 1

Expand the square (see special products of binomials)

1 + 2 \sin x \cos x = 1

Replace $\sin^2 x + \cos^2 x$ with 1.

2 \sin x \cos x = 0

Subtract 1 from both sides.

\sin x \cos x = 0

Divide both sides by 2.

Set each factor equal to 0: $\sin x = 0$ gives $x = 0, \pi$ , and $\cos x = 0$ gives $x = \frac{\pi}{2}, \frac{3\pi}{2}$

Squaring can add false answers (see extraneous solutions), so check each one in the original: $x = 0$ gives $0 + 1 = 1$ , which works

Check $x = \frac{\pi}{2}$ : $1 + 0 = 1$ , which works

Check $x = \pi$ : $0 + (-1) = -1$ , which is not 1, so reject it

Check $x = \frac{3\pi}{2}$ : $-1 + 0 = -1$ , which is not 1, so reject it

Answer: $x = 0, \frac{\pi}{2}$

Example 5: Word problem with a kicked ball

A ball is kicked at 64 feet per second at an angle $\theta$ above the ground. It lands $R = 256 \sin \theta \cos \theta$ feet away. For $0^\circ \lt \theta \lt 90^\circ$ , at what angles does it land $64\sqrt{3}$ feet away (about 111 feet)?

256 \sin \theta \cos \theta = 64\sqrt{3}

Set the distance equal to $64\sqrt{3}$ .

128 (2 \sin \theta \cos \theta) = 64\sqrt{3}

Write 256 as $128 \cdot 2$ .

128 \sin 2\theta = 64\sqrt{3}

Replace $2 \sin \theta \cos \theta$ with $\sin 2\theta$ .

\sin 2\theta = \frac{\sqrt{3}}{2}

Divide both sides by 128.

Since $0^\circ \lt \theta \lt 90^\circ$ , the double angle is in $0^\circ \lt 2\theta \lt 180^\circ$ . Sine is $\frac{\sqrt{3}}{2}$ at $2\theta = 60^\circ$ and $2\theta = 120^\circ$

Divide each by 2: $\theta = 30^\circ$ or $\theta = 60^\circ$

Answer: $30^\circ$ or $60^\circ$ . A low kick and a high kick land in the same spot, because $\sin 60^\circ = \sin 120^\circ$ .

Common Mistakes

Mistake	Example of the error	Fix
Dividing by a trig function	Dividing $2 \sin x \cos x = \cos x$ by $\cos x$	Move all terms to one side and factor out $\cos x$ , so you keep $\cos x = 0$ .
Picking the wrong $\cos 2x$ formula	Using $\cos^2 x - \sin^2 x$ when the rest is in $\cos x$	Choose the form that matches: $2\cos^2 x - 1$ for cosine, $1 - 2\sin^2 x$ for sine.
Sign error in the Pythagorean swap	$\cos^2 x = \sin^2 x - 1$	It is $\cos^2 x = 1 - \sin^2 x$ .
Not checking after squaring	Keeping $x = \pi$ for $\sin x + \cos x = 1$	Plug every answer into the original equation.
Forgetting one factor	Solving only $\sin x = \frac{1}{2}$ from $(2\sin x - 1)(\sin x - 1) = 0$	Set every factor equal to 0.

Related Methods

Sum and difference formulas

Some equations have angles like $x + \frac{\pi}{6}$ . Expand them with the sum and difference formulas first. For example, $\sin(x + \frac{\pi}{6}) + \sin(x - \frac{\pi}{6})$ simplifies to $\sqrt{3} \sin x$ .

Use a multiple angle instead

Sometimes it is faster to go the other way. In $\sin x \cos x = \frac{1}{4}$ , multiply by 2 to get $\sin 2x = \frac{1}{2}$ . Then solve it as in trig equations with multiple angles.

Check with a graph

Graph each side as its own function, as in the picture above. The number of crossing points on $[0, 2\pi)$ tells you how many answers to expect. This is also how you spot an extraneous answer.

Practice Problems

Solve $2\cos^2 x - \sin x - 1 = 0$ on $[0, 2\pi)$ .
Solve $\sec^2 x - \tan x - 1 = 0$ on $[0, 2\pi)$ .
Solve $\sin 2x = \sin x$ on $[0, 2\pi)$ .
Solve $\cos 2x = \cos x$ on $[0, 2\pi)$ .
Solve $\cos 2x + \sin x = 0$ on $[0, 2\pi)$ .
Solve $\tan x + \cot x = 2$ on $[0, 2\pi)$ .
Solve $\sin^2 x - \cos^2 x = \frac{1}{2}$ on $[0, 2\pi)$ with a Pythagorean identity.
Solve $\cos x - \sin x = 1$ on $[0, 2\pi)$ .
Solve $\sin(x + \frac{\pi}{6}) + \sin(x - \frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ on $[0, 2\pi)$ .
A ball kicked at 48 feet per second lands $R = 144 \sin \theta \cos \theta$ feet away. For $0^\circ \lt \theta \lt 90^\circ$ , at what angles does it land 36 feet away?

Show answers

2(1 - \sin^2 x) - \sin x - 1 = 0

Replace $\cos^2 x$ with $1 - \sin^2 x$ .

2 - 2\sin^2 x - \sin x - 1 = 0

Distribute the 2.

-2\sin^2 x - \sin x + 1 = 0

Combine the constants.

2\sin^2 x + \sin x - 1 = 0

Multiply both sides by −1.

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

Factor.

$2 \sin x - 1 = 0$ gives $\sin x = \frac{1}{2}$ , so $x = \frac{\pi}{6}, \frac{5\pi}{6}$
$\sin x + 1 = 0$ gives $\sin x = -1$ , so $x = \frac{3\pi}{2}$
Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$

1 + \tan^2 x - \tan x - 1 = 0

Replace $\sec^2 x$ with $1 + \tan^2 x$ .

\tan^2 x - \tan x = 0

Combine the constants.

\tan x (\tan x - 1) = 0

Factor out $\tan x$ .

$\tan x = 0$ gives $x = 0, \pi$
$\tan x - 1 = 0$ gives $\tan x = 1$ , so $x = \frac{\pi}{4}, \frac{5\pi}{4}$
Answer: $x = 0, \frac{\pi}{4}, \pi, \frac{5\pi}{4}$

2 \sin x \cos x = \sin x

Replace $\sin 2x$ with $2 \sin x \cos x$ .

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

Subtract $\sin x$ from both sides.

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

Factor out $\sin x$ .

$\sin x = 0$ gives $x = 0, \pi$
$2 \cos x - 1 = 0$ gives $\cos x = \frac{1}{2}$ , so $x = \frac{\pi}{3}, \frac{5\pi}{3}$
Answer: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$

2\cos^2 x - 1 = \cos x

Replace $\cos 2x$ with $2\cos^2 x - 1$ .

2\cos^2 x - \cos x - 1 = 0

Subtract $\cos x$ from both sides.

(2 \cos x + 1)(\cos x - 1) = 0

Factor.

$2 \cos x + 1 = 0$ gives $\cos x = -\frac{1}{2}$ , so $x = \frac{2\pi}{3}, \frac{4\pi}{3}$
$\cos x - 1 = 0$ gives $\cos x = 1$ , so $x = 0$
Answer: $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$

1 - 2\sin^2 x + \sin x = 0

Replace $\cos 2x$ with $1 - 2\sin^2 x$ .

2\sin^2 x - \sin x - 1 = 0

Multiply both sides by −1.

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

Factor.

$2 \sin x + 1 = 0$ gives $\sin x = -\frac{1}{2}$ , so $x = \frac{7\pi}{6}, \frac{11\pi}{6}$
$\sin x - 1 = 0$ gives $\sin x = 1$ , so $x = \frac{\pi}{2}$
Answer: $x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$

\tan x + \frac{1}{\tan x} = 2

Replace $\cot x$ with $\frac{1}{\tan x}$ .

\tan^2 x + 1 = 2 \tan x

Multiply both sides by $\tan x$ .

\tan^2 x - 2 \tan x + 1 = 0

Subtract $2 \tan x$ from both sides.

(\tan x - 1)^2 = 0

Factor the perfect square.

\tan x = 1

Set the factor equal to 0.

Answer: $x = \frac{\pi}{4}, \frac{5\pi}{4}$

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

Replace $\cos^2 x$ with $1 - \sin^2 x$ .

2\sin^2 x - 1 = \frac{1}{2}

Remove the parentheses.

2\sin^2 x = \frac{3}{2}

Add 1 to both sides.

\sin^2 x = \frac{3}{4}

Divide both sides by 2.

\sin x = \pm\frac{\sqrt{3}}{2}

Take the square root of both sides.

$\sin x = \frac{\sqrt{3}}{2}$ gives $x = \frac{\pi}{3}, \frac{2\pi}{3}$ ; $\sin x = -\frac{\sqrt{3}}{2}$ gives $x = \frac{4\pi}{3}, \frac{5\pi}{3}$
Answer: $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$

\cos^2 x - 2 \sin x \cos x + \sin^2 x = 1

Square both sides.

1 - 2 \sin x \cos x = 1

Replace $\sin^2 x + \cos^2 x$ with 1.

-2 \sin x \cos x = 0

Subtract 1 from both sides.

\sin x \cos x = 0

Divide both sides by −2.

x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}

List the angles where sine or cosine is 0.

Check: $x = 0$ gives $1 - 0 = 1$ (works); $\frac{\pi}{2}$ gives $0 - 1 = -1$ (reject); $\pi$ gives $-1 - 0 = -1$ (reject); $\frac{3\pi}{2}$ gives $0 - (-1) = 1$ (works)
Answer: $x = 0, \frac{3\pi}{2}$

\sin x \cos \frac{\pi}{6} + \cos x \sin \frac{\pi}{6} + \sin x \cos \frac{\pi}{6} - \cos x \sin \frac{\pi}{6}

Expand with the sum and difference formulas.

2 \sin x \cos \frac{\pi}{6}

The $\cos x$ terms cancel.

2 \sin x \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \sin x

Use $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ .

\sqrt{3} \sin x = \frac{\sqrt{3}}{2}

Set it equal.

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

Divide both sides by $\sqrt{3}$ .

Answer: $x = \frac{\pi}{6}, \frac{5\pi}{6}$

10.

144 \sin \theta \cos \theta = 36

Set the distance to 36.

72 (2 \sin \theta \cos \theta) = 36

Write 144 as $72 \cdot 2$ .

72 \sin 2\theta = 36

Replace $2 \sin \theta \cos \theta$ with $\sin 2\theta$ .

\sin 2\theta = \frac{1}{2}

Divide both sides by 72.

For $0^\circ \lt 2\theta \lt 180^\circ$ : $2\theta = 30^\circ$ or $2\theta = 150^\circ$
Divide each by 2: $\theta = 15^\circ$ or $\theta = 75^\circ$
Answer: $15^\circ$ or $75^\circ$

FAQ

How do you solve trig equations using identities?

Use an identity to rewrite the equation with one trig function of one angle. Then move everything to one side, factor and solve each factor on the interval.

Which identity should I use first?

Look at what doesn’t match. Mixed functions call for a Pythagorean or reciprocal identity. Mixed angles, like $2x$ and $x$ , call for a double-angle formula.

Why do I need to check answers after squaring?

Squaring makes negative and positive values look the same, so it can add answers that don’t work. Plug each one into the original equation and drop any that fail.

Can I divide both sides by sin x or cos x?

No. Dividing throws away the answers where that function equals 0. Factor it out instead and set each factor equal to 0.

Summary

Use an identity to get one trig function of one angle.
Pythagorean identities fix mixed functions; double-angle formulas fix mixed angles.
Move every term to one side, factor, and set each factor equal to 0.
Never divide by a trig function, because you lose answers.
If you square both sides, check every answer in the original equation.

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