1.8.7 Solving trigonometric equations

What you'll learn

How to solve equations involving $sin⁡x\sin x$ , $cos⁡x\cos x$ and $tan⁡x\tan x$ in a given interval.
How symmetry and periodicity produce additional solutions.
How to solve equations that are quadratic in a trigonometric function.
How to handle multiples of the unknown angle, such as $sin⁡2x\sin 2x$ or $cos⁡3x\cos 3x$ .

Before you begin

You should be comfortable with exact trigonometric values, rearranging equations and solving quadratic equations.

Always check whether the question is working in degrees or radians. An interval containing $π\pi$ uses radians, while an interval containing a degree symbol uses degrees. Make sure your calculator is in the correct mode.

Periodicity and symmetry

Periodicity

A trigonometric function is periodic if its values repeat after a fixed interval. This repeating interval is called the period.

Definition

Periods of the trigonometric functions

The functions $sin⁡x\sin x$ and $cos⁡x\cos x$ have period $2π2\pi$ radians, or 360°.
The function $tan⁡x\tan x$ has period $π\pi$ radians, or 180°.

Therefore,

sin⁡(x+2π)=sin⁡x,cos⁡(x+2π)=cos⁡x,tan⁡(x+π)=tan⁡x.\begin{aligned} \sin(x+2\pi)&=\sin x,\\ \cos(x+2\pi)&=\cos x,\\ \tan(x+\pi)&=\tan x. \end{aligned}

Periodicity means that a trigonometric equation will usually have infinitely many solutions overall. A stated interval tells you which of these solutions to include.

Symmetry

Your calculator gives one principal value, meaning its standard inverse-trigonometric answer. This is not necessarily the only solution in the interval.

For an angle $α\alpha$ :

x=α+π,\begin{aligned} \sin x=\sin\alpha &\Rightarrow x=\alpha \text{ or } x=\pi-\alpha,\\ \cos x=\cos\alpha &\Rightarrow x=\alpha \text{ or } x=2\pi-\alpha,\\ \tan x=\tan\alpha &\Rightarrow x=\alpha \text{ or } x=\alpha+\pi, \end{aligned}

when searching through one full turn from $0$ to $2π2\pi$ .

Key Idea

One calculator answer is not enough

Use the first answer to find a reference angle, then use the signs, symmetry and period of the function to locate every solution in the required interval.

The graph below shows why $sin⁡x=12\sin x=\frac12$ has two solutions between $0$ and $2π2\pi$ : the sine curve meets the horizontal line twice.

Graph of y = sin x and y = 1/2 showing intersections at pi/6 and 5pi/6

Solving simple trigonometric equations

A simple trigonometric equation can be rearranged into one of the forms

sin⁡x=a,cos⁡x=a,tan⁡x=a.\sin x=a,\qquad \cos x=a,\qquad \tan x=a.

A reliable method is:

Rearrange to isolate the trigonometric function.
Find one angle using the appropriate inverse function.
Use symmetry and periodicity to find the other angles.
Keep only the solutions in the stated interval.

Example

Solving a sine equation

Solve $2sin⁡x−1=02\sin x-1=0$ for $0≤x<2π0\le x<2\pi$ .

Isolate the sine term:
$2sin⁡x−1=0⇒sin⁡x=12.2\sin x-1=0 \Rightarrow \sin x=\frac12.$
Find the first solution using the exact value $sin⁡π6=12\sin\frac{\pi}{6}=\frac12$ , so the reference angle is $π6\frac{\pi}{6}$ .
Sine is positive in the first and second quadrants. The second solution is
$π−π6=5π6.\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$
Both angles lie in the interval, so
$5π6.x=\frac{\pi}{6},\ \frac{5\pi}{6}.$

Common Mistake

Missing the second solution

Writing only $x=sin⁡−1(12)x=\sin^{-1}\left(\frac12\right)$ misses the second intersection of the sine graph with the line $y=12y=\frac12$ .

Negative trigonometric values

The sign tells you which quadrants contain solutions:

Sine is positive in quadrants 1 and 2.
Cosine is positive in quadrants 1 and 4.
Tangent is positive in quadrants 1 and 3.

For a negative value, use the remaining two quadrants.

Tip

Use a positive reference angle

When the trigonometric value is negative, find the acute reference angle from its positive magnitude. Then place that angle in the quadrants where the required function is negative.

Example

Solving a negative cosine equation

Solve $cos⁡x=−32\cos x=-\frac{\sqrt3}{2}$ for $0≤x<2π0\le x<2\pi$ .

The reference angle is $π6\frac{\pi}{6}$ because $cos⁡π6=32\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$ .
Cosine is negative in quadrants 2 and 3, giving
$x=π−π6orx=π+π6.x=\pi-\frac{\pi}{6} \quad\text{or}\quad x=\pi+\frac{\pi}{6}.$
Therefore,
$7π6.x=\frac{5\pi}{6},\ \frac{7\pi}{6}.$

Checking whether solutions are possible

For every real angle,

−1≤sin⁡x≤1and−1≤cos⁡x≤1.-1\le\sin x\le1 \qquad\text{and}\qquad -1\le\cos x\le1.

Therefore, equations such as $sin⁡x=1.4\sin x=1.4$ and $cos⁡x=−2\cos x=-2$ have no real solutions.

Tangent is not restricted to this range, so $tan⁡x=a\tan x=a$ can have solutions for any real value of $a$ .

Common Mistake

Reject impossible roots

When algebra produces a value outside the interval from negative one to one for $sin⁡x\sin x$ or $cos⁡x\cos x$ , reject that value before trying to find an angle.

Quadratic trigonometric equations

A quadratic trigonometric equation is quadratic in one trigonometric expression. For example,

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

Treat the trigonometric function temporarily as one algebraic variable. You can factorise or use the quadratic formula, then solve the resulting simple trigonometric equations.

Example

Solving a quadratic equation in sine

Solve $2\sin^2x-\sin x-1=0$ for $0≤x<2π0\le x<2\pi$ .

Let $u=sin⁡xu=\sin x$ . The equation becomes
$2u^2-u-1=0.$
Factorise:
$(2 u + 1) (u - 1) = 0,$
so
$u=−12oru=1.u=-\frac12 \quad\text{or}\quad u=1.$
Substitute back:
$sin⁡x=−12orsin⁡x=1.\sin x=-\frac12 \quad\text{or}\quad \sin x=1.$
Sine equals $−12-\frac12$ in quadrants 3 and 4, giving $x=7π6x=\frac{7\pi}{6}$ and $x=11π6x=\frac{11\pi}{6}$ . Sine equals 1 at $x=π2x=\frac{\pi}{2}$ .
Therefore,
$11π6.x=\frac{\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}.$

Common Mistake

Losing one branch

After factorising, solve every resulting equation. It is easy to solve one factor and accidentally ignore the other.

The same technique applies to quadratics in cosine or tangent. Remember that possible cosine roots must also lie between negative one and one.

Equations involving multiples of the angle

An equation such as $sin⁡2x=12\sin 2x=\frac12$ involves a multiple angle because the input to sine is $2 x$ , not $x$ .

Introduce a temporary angle, such as $θ=2x\theta=2x$ . Most importantly, transform the interval as well.

Key Idea

Transform the interval

If $a≤x≤ba\le x\le b$ and $θ=kx\theta=kx$ for positive $k$ , then search over $ka≤θ≤kbka\le\theta\le kb$ . This prevents you from missing solutions.

Example

Solving an equation involving a double angle

Solve $sin⁡2x=12\sin 2x=\frac12$ for $0≤x<2π0\le x<2\pi$ .

Let $θ=2x\theta=2x$ . Since $0≤x<2π0\le x<2\pi$ , the transformed interval is
$0≤θ<4π.0\le\theta<4\pi.$
Solve $sin⁡θ=12\sin\theta=\frac12$ . In the first cycle, the solutions are
$5π6.\theta=\frac{\pi}{6},\ \frac{5\pi}{6}.$
The interval extends through a second cycle, so add the sine period $2π2\pi$ :
$17π6.\theta=\frac{13\pi}{6},\ \frac{17\pi}{6}.$
Divide every value by 2 because $θ=2x\theta=2x$ :
$17π12.x=\frac{\pi}{12},\ \frac{5\pi}{12},\ \frac{13\pi}{12},\ \frac{17\pi}{12}.$

Common Mistake

Using the original interval

If you solve $sin⁡2x=12\sin 2x=\frac12$ only over $0≤2x<2π0\le2x<2\pi$ , you search through only half of the required range and miss valid solutions.

Combining quadratics and multiple angles

Some questions combine both techniques. First solve the quadratic in the complete trigonometric expression, then solve each resulting multiple-angle equation over a transformed interval.

Example

Combining a quadratic with a multiple angle

Solve $2\cos^2 2x-\cos 2x-1=0$ for $0≤x<π0\le x<\pi$ .

Let $u=cos⁡2xu=\cos 2x$ and factorise:
$2u^2-u-1=(2u+1)(u-1)=0.$
Hence $cos⁡2x=−12\cos 2x=-\frac12$ or $cos⁡2x=1\cos 2x=1$ .
Let $θ=2x\theta=2x$ . The interval becomes $0≤θ<2π0\le\theta<2\pi$ .
Within this interval,
$4π3,\cos\theta=-\frac12 \Rightarrow \theta=\frac{2\pi}{3},\ \frac{4\pi}{3},$
and
$cos⁡θ=1⇒θ=0.\cos\theta=1 \Rightarrow \theta=0.$
Divide each value by 2:
$2π3.x=0,\ \frac{\pi}{3},\ \frac{2\pi}{3}.$

Exam technique

In the exam

Isolate the trigonometric function, or factorise first if the equation is quadratic.
Write down every algebraic branch and reject impossible sine or cosine values.
For $sin⁡kx\sin kx$ , $cos⁡kx\cos kx$ or $tan⁡kx\tan kx$ , transform the interval before finding solutions.
Use symmetry and periodicity to find every angle, then check each endpoint carefully.
Give exact answers involving $π\pi$ , fractions or surds unless the question requests decimals.

Self review

Check yourself

Can you solve $tan⁡x=−1\tan x=-1$ for $0≤x<2π0\le x<2\pi$ ?
How would you solve $2\cos^2x+\cos x-1=0$ over one full turn?
What interval should you use for $θ=3x\theta=3x$ if $−π≤x≤π-\pi\le x\le\pi$ ?

What you'll learn

How to solve equations involving $sin⁡x\sin x$ , $cos⁡x\cos x$ and $tan⁡x\tan x$ in a given interval.
How symmetry and periodicity produce additional solutions.
How to solve equations that are quadratic in a trigonometric function.
How to handle multiples of the unknown angle, such as $sin⁡2x\sin 2x$ or $cos⁡3x\cos 3x$ .

Before you begin

You should be comfortable with exact trigonometric values, rearranging equations and solving quadratic equations.

Periodicity and symmetry

Periodicity

A trigonometric function is periodic if its values repeat after a fixed interval. This repeating interval is called the period.

Definition

Periods of the trigonometric functions

The functions $sin⁡x\sin x$ and $cos⁡x\cos x$ have period $2π2\pi$ radians, or 360°.
The function $tan⁡x\tan x$ has period $π\pi$ radians, or 180°.

Therefore,

sin⁡(x+2π)=sin⁡x,cos⁡(x+2π)=cos⁡x,tan⁡(x+π)=tan⁡x.\begin{aligned} \sin(x+2\pi)&=\sin x,\\ \cos(x+2\pi)&=\cos x,\\ \tan(x+\pi)&=\tan x. \end{aligned}

Periodicity means that a trigonometric equation will usually have infinitely many solutions overall. A stated interval tells you which of these solutions to include.

Symmetry

Your calculator gives one principal value, meaning its standard inverse-trigonometric answer. This is not necessarily the only solution in the interval.

For an angle $α\alpha$ :

x=α+π,\begin{aligned} \sin x=\sin\alpha &\Rightarrow x=\alpha \text{ or } x=\pi-\alpha,\\ \cos x=\cos\alpha &\Rightarrow x=\alpha \text{ or } x=2\pi-\alpha,\\ \tan x=\tan\alpha &\Rightarrow x=\alpha \text{ or } x=\alpha+\pi, \end{aligned}

when searching through one full turn from $0$ to $2π2\pi$ .

Key Idea

One calculator answer is not enough

Use the first answer to find a reference angle, then use the signs, symmetry and period of the function to locate every solution in the required interval.

The graph below shows why $sin⁡x=12\sin x=\frac12$ has two solutions between $0$ and $2π2\pi$ : the sine curve meets the horizontal line twice.

Graph of y = sin x and y = 1/2 showing intersections at pi/6 and 5pi/6

Solving simple trigonometric equations

A simple trigonometric equation can be rearranged into one of the forms

sin⁡x=a,cos⁡x=a,tan⁡x=a.\sin x=a,\qquad \cos x=a,\qquad \tan x=a.

A reliable method is:

Rearrange to isolate the trigonometric function.
Find one angle using the appropriate inverse function.
Use symmetry and periodicity to find the other angles.
Keep only the solutions in the stated interval.

Example

Solving a sine equation

Solve $2sin⁡x−1=02\sin x-1=0$ for $0≤x<2π0\le x<2\pi$ .

Isolate the sine term:
$2sin⁡x−1=0⇒sin⁡x=12.2\sin x-1=0 \Rightarrow \sin x=\frac12.$
Find the first solution using the exact value $sin⁡π6=12\sin\frac{\pi}{6}=\frac12$ , so the reference angle is $π6\frac{\pi}{6}$ .
Sine is positive in the first and second quadrants. The second solution is
$π−π6=5π6.\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$
Both angles lie in the interval, so
$5π6.x=\frac{\pi}{6},\ \frac{5\pi}{6}.$

Common Mistake

Missing the second solution

Writing only $x=sin⁡−1(12)x=\sin^{-1}\left(\frac12\right)$ misses the second intersection of the sine graph with the line $y=12y=\frac12$ .

Negative trigonometric values

The sign tells you which quadrants contain solutions:

Sine is positive in quadrants 1 and 2.
Cosine is positive in quadrants 1 and 4.
Tangent is positive in quadrants 1 and 3.

For a negative value, use the remaining two quadrants.

Tip

Use a positive reference angle

When the trigonometric value is negative, find the acute reference angle from its positive magnitude. Then place that angle in the quadrants where the required function is negative.

Example

Solving a negative cosine equation

Solve $cos⁡x=−32\cos x=-\frac{\sqrt3}{2}$ for $0≤x<2π0\le x<2\pi$ .

The reference angle is $π6\frac{\pi}{6}$ because $cos⁡π6=32\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$ .
Cosine is negative in quadrants 2 and 3, giving
$x=π−π6orx=π+π6.x=\pi-\frac{\pi}{6} \quad\text{or}\quad x=\pi+\frac{\pi}{6}.$
Therefore,
$7π6.x=\frac{5\pi}{6},\ \frac{7\pi}{6}.$

Checking whether solutions are possible

For every real angle,

−1≤sin⁡x≤1and−1≤cos⁡x≤1.-1\le\sin x\le1 \qquad\text{and}\qquad -1\le\cos x\le1.

Therefore, equations such as $sin⁡x=1.4\sin x=1.4$ and $cos⁡x=−2\cos x=-2$ have no real solutions.

Tangent is not restricted to this range, so $tan⁡x=a\tan x=a$ can have solutions for any real value of $a$ .

Common Mistake

Reject impossible roots

When algebra produces a value outside the interval from negative one to one for $sin⁡x\sin x$ or $cos⁡x\cos x$ , reject that value before trying to find an angle.

Quadratic trigonometric equations

A quadratic trigonometric equation is quadratic in one trigonometric expression. For example,

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

Treat the trigonometric function temporarily as one algebraic variable. You can factorise or use the quadratic formula, then solve the resulting simple trigonometric equations.

Example

Solving a quadratic equation in sine

Solve $2\sin^2x-\sin x-1=0$ for $0≤x<2π0\le x<2\pi$ .

Let $u=sin⁡xu=\sin x$ . The equation becomes
$2u^2-u-1=0.$
Factorise:
$(2 u + 1) (u - 1) = 0,$
so
$u=−12oru=1.u=-\frac12 \quad\text{or}\quad u=1.$
Substitute back:
$sin⁡x=−12orsin⁡x=1.\sin x=-\frac12 \quad\text{or}\quad \sin x=1.$
Sine equals $−12-\frac12$ in quadrants 3 and 4, giving $x=7π6x=\frac{7\pi}{6}$ and $x=11π6x=\frac{11\pi}{6}$ . Sine equals 1 at $x=π2x=\frac{\pi}{2}$ .
Therefore,
$11π6.x=\frac{\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}.$

Common Mistake

Losing one branch

After factorising, solve every resulting equation. It is easy to solve one factor and accidentally ignore the other.

The same technique applies to quadratics in cosine or tangent. Remember that possible cosine roots must also lie between negative one and one.

Equations involving multiples of the angle

An equation such as $sin⁡2x=12\sin 2x=\frac12$ involves a multiple angle because the input to sine is $2 x$ , not $x$ .

Introduce a temporary angle, such as $θ=2x\theta=2x$ . Most importantly, transform the interval as well.

Key Idea

Transform the interval

If $a≤x≤ba\le x\le b$ and $θ=kx\theta=kx$ for positive $k$ , then search over $ka≤θ≤kbka\le\theta\le kb$ . This prevents you from missing solutions.

Example

Solving an equation involving a double angle

Solve $sin⁡2x=12\sin 2x=\frac12$ for $0≤x<2π0\le x<2\pi$ .

Let $θ=2x\theta=2x$ . Since $0≤x<2π0\le x<2\pi$ , the transformed interval is
$0≤θ<4π.0\le\theta<4\pi.$
Solve $sin⁡θ=12\sin\theta=\frac12$ . In the first cycle, the solutions are
$5π6.\theta=\frac{\pi}{6},\ \frac{5\pi}{6}.$
The interval extends through a second cycle, so add the sine period $2π2\pi$ :
$17π6.\theta=\frac{13\pi}{6},\ \frac{17\pi}{6}.$
Divide every value by 2 because $θ=2x\theta=2x$ :
$17π12.x=\frac{\pi}{12},\ \frac{5\pi}{12},\ \frac{13\pi}{12},\ \frac{17\pi}{12}.$

Common Mistake

Using the original interval

If you solve $sin⁡2x=12\sin 2x=\frac12$ only over $0≤2x<2π0\le2x<2\pi$ , you search through only half of the required range and miss valid solutions.

Combining quadratics and multiple angles

Some questions combine both techniques. First solve the quadratic in the complete trigonometric expression, then solve each resulting multiple-angle equation over a transformed interval.

Example

Combining a quadratic with a multiple angle

Solve $2\cos^2 2x-\cos 2x-1=0$ for $0≤x<π0\le x<\pi$ .

Let $u=cos⁡2xu=\cos 2x$ and factorise:
$2u^2-u-1=(2u+1)(u-1)=0.$
Hence $cos⁡2x=−12\cos 2x=-\frac12$ or $cos⁡2x=1\cos 2x=1$ .
Let $θ=2x\theta=2x$ . The interval becomes $0≤θ<2π0\le\theta<2\pi$ .
Within this interval,
$4π3,\cos\theta=-\frac12 \Rightarrow \theta=\frac{2\pi}{3},\ \frac{4\pi}{3},$
and
$cos⁡θ=1⇒θ=0.\cos\theta=1 \Rightarrow \theta=0.$
Divide each value by 2:
$2π3.x=0,\ \frac{\pi}{3},\ \frac{2\pi}{3}.$

Exam technique

In the exam

Isolate the trigonometric function, or factorise first if the equation is quadratic.
Write down every algebraic branch and reject impossible sine or cosine values.
For $sin⁡kx\sin kx$ , $cos⁡kx\cos kx$ or $tan⁡kx\tan kx$ , transform the interval before finding solutions.
Use symmetry and periodicity to find every angle, then check each endpoint carefully.
Give exact answers involving $π\pi$ , fractions or surds unless the question requests decimals.

Self review

Check yourself

Can you solve $tan⁡x=−1\tan x=-1$ for $0≤x<2π0\le x<2\pi$ ?
How would you solve $2\cos^2x+\cos x-1=0$ over one full turn?
What interval should you use for $θ=3x\theta=3x$ if $−π≤x≤π-\pi\le x\le\pi$ ?

What you'll learn

Before you begin

Periodicity and symmetry

Periodicity

Symmetry

Solving simple trigonometric equations

Negative trigonometric values

Checking whether solutions are possible

Quadratic trigonometric equations

Equations involving multiples of the angle

Combining quadratics and multiple angles

Review 1.8.7 Solving trigonometric equations by teaching Genie

1.8.7 Solving trigonometric equations

What you'll learn

Before you begin

Periodicity and symmetry

Periodicity

Symmetry

Solving simple trigonometric equations

Negative trigonometric values

Checking whether solutions are possible

Quadratic trigonometric equations

Equations involving multiples of the angle

Combining quadratics and multiple angles

Review 1.8.7 Solving trigonometric equations by teaching Genie

1.8.7 Solving trigonometric equations

What you'll learn

Before you begin

Periodicity and symmetry

Periodicity

Symmetry

Solving simple trigonometric equations

Negative trigonometric values

Checking whether solutions are possible

Quadratic trigonometric equations

Equations involving multiples of the angle

Combining quadratics and multiple angles

Review 1.8.7 Solving trigonometric equations by teaching Genie

1.8.7 Solving trigonometric equations Revision Guide

1.8.7 Solving trigonometric equations

What you'll learn

Before you begin

Periodicity and symmetry

Periodicity

Symmetry

Solving simple trigonometric equations

Negative trigonometric values

Checking whether solutions are possible

Quadratic trigonometric equations

Equations involving multiples of the angle

Combining quadratics and multiple angles

Review 1.8.7 Solving trigonometric equations by teaching Genie

1.8.7 Solving trigonometric equations Revision Guide