10.4 Solving Trigonometric Equations

What you'll learn

How to find all solutions to trigonometric equations in a given interval.
How to handle shifted or stretched angles such as $2x+10∘2x+10^\circ$ .
How to solve equations involving $sin^2 x$ , $cos^2 x$ or $tan^2 x$ .
How to use identities and factorisation without losing valid answers.

The big idea: one trig equation can have several answers

A trigonometric equation is an equation involving a trig function such as $sin⁡x\sin x$ , $cos⁡x\cos x$ or $tan⁡x\tan x$ .

In this topic, angles are measured in degrees, so make sure your calculator is in degree mode. The key challenge is that trig graphs repeat, so one calculator answer is usually not enough.

Definition

Solution interval

A solution interval tells you the range of angles you are allowed to give as answers. For example, $\le x < 360^\circ$ means include 0°, include everything up to 360°, but do not include 360° itself.

A graph helps you see why there can be more than one answer. Solving $sin⁡(x−30∘)=0.3\sin(x-30^\circ)=0.3$ means finding where the sine curve meets the horizontal line $y = 0.3$ .

The shifted sine graph intersects the line y=0.3 more than once, showing why a trigonometric equation can have several solutions.

Graph of y = sin(x - 30°) and y = 0.3 showing two intersections

Key Idea

The calculator gives a starting point

Your calculator gives a principal value, which is the first angle it finds. You must then use symmetry and periodicity to find all the other angles in the required interval.

A reliable routine for shifted sine and cosine equations

For equations like $cos⁡(2x+10∘)=0.4\cos(2x+10^\circ)=0.4$ , the expression inside the trig function is not just $x$ . A good method is to temporarily replace the inside angle with a new letter.

Definition

Period

The period of a trig function is how long it takes before the graph repeats. Sine and cosine repeat every 360°, while tangent repeats every 180°.

Example

Solving a shifted cosine equation

Solve $cos⁡(2x+10∘)=0.4\cos(2x+10^\circ)=0.4$ for $\le x < 180^\circ$ . Give your answers to one decimal place.

Let the inside angle be $u$ :
$2x+10^\circ$
Convert the interval for $x$ into an interval for $u$ . If $\le x < 180^\circ$ , then:

$The substitution u=2x+10^\circ stretches and shifts the original x-interval to the corresponding u-interval.$

$$
10^\circ \le u < 370^\circ
$$

3. Use your calculator to find the principal value:

$$
\arccos(0.4)=66.4218\ldots^\circ
$$

4. Cosine is positive in quadrants I and IV, so within $10∘≤u<370∘10^\circ \le u < 370^\circ$ :

Cosine is positive in quadrants I and IV, giving two possible u-angles in the interval.

$$
u = 66.4218\ldots^\circ,\quad 293.5781\ldots^\circ
$$

5. Convert back to $x$ using $u=2x+10∘u=2x+10^\circ$ :

$$
\begin{aligned}
x &= \frac{66.4218\ldots-10}{2}=28.2109\ldots^\circ \\
x &= \frac{293.5781\ldots-10}{2}=141.7890\ldots^\circ
\end{aligned}
$$

6. Round to one decimal place:

$$
x = 28.2^\circ,\ 141.8^\circ
$$

Tip

Work with the inside angle first

If the equation contains $sin⁡(3θ−20∘)\sin(3\theta-20^\circ)$ , solve for the whole angle $3θ−20∘3\theta-20^\circ$ first. Only divide by 3 and add 20° at the end.

Tangent equations: remember the shorter period

Tangent behaves differently from sine and cosine because $tan⁡x\tan x$ repeats every 180°. This often makes tangent equations quicker, but it is easy to miss a solution if you only use the calculator answer.

Example

Solving a shifted tangent equation

Solve $tan⁡(θ+40∘)=−1.8\tan(\theta+40^\circ)=-1.8$ for $−180∘≤θ<180∘-180^\circ \le \theta < 180^\circ$ . Give your answers to one decimal place.

Let the inside angle be $u$ :
$u=θ+40∘u=\theta+40^\circ$
Convert the interval. If $−180∘≤θ<180∘-180^\circ \le \theta < 180^\circ$ , then:
$−140∘≤u<220∘-140^\circ \le u < 220^\circ$
Find one calculator solution:
$arctan⁡(−1.8)=−60.9453…∘\arctan(-1.8)=-60.9453\ldots^\circ$
Since tangent has period 180°, add or subtract 180° until you have all values in the interval:

$Tangent solutions repeat every 180^\circ, so the calculator solution must be shifted by multiples of 180^\circ within the allowed interval.$

$$
u=-60.9453\ldots^\circ,\quad 119.0546\ldots^\circ
$$

5. Convert back to $θ\theta$ using $θ=u−40∘\theta=u-40^\circ$ :

$$
\theta=-100.9453\ldots^\circ,\quad 79.0546\ldots^\circ
$$

6. Round to one decimal place:

$$
\theta=-100.9^\circ,\ 79.1^\circ
$$

Squared trig functions

A squared trig function means the whole trig value is squared. For example, $tan^2 x$ means $tan x)^2$ , not $tan(x^2)$ .

When you square-root both sides, remember the positive and negative possibilities.

Example

Solving a squared tangent equation

Solve $tan^2 x=5$ for $\le x < 360^\circ$ . Give your answers to one decimal place.

Square-root both sides:
$tan⁡x=±5\tan x=\pm\sqrt{5}$
Find the reference angle:

The four tangent solutions come from using the same reference angle in all four quadrants after taking both positive and negative square roots.

$$
\arctan(\sqrt{5})=65.9051\ldots^\circ
$$

3. For $tan⁡x=5\tan x=\sqrt{5}$ , tangent is positive in quadrants I and III:

$$
x=65.9051\ldots^\circ,\quad 245.9051\ldots^\circ
$$

4. For $tan⁡x=−5\tan x=-\sqrt{5}$ , tangent is negative in quadrants II and IV:

$$
x=114.0948\ldots^\circ,\quad 294.0948\ldots^\circ
$$

5. Round to one decimal place:

$$
x=65.9^\circ,\ 114.1^\circ,\ 245.9^\circ,\ 294.1^\circ
$$

Common Mistake

Forgetting the negative square root

From $tan^2 x=5$ , do not only write $tan⁡x=5\tan x=\sqrt{5}$ . You also need $tan⁡x=−5\tan x=-\sqrt{5}$ .

Using identities to make quadratics

An identity is an equation that is true for all allowed values of the variable. The most useful identity here is:

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

So:

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

and:

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

A quadratic equation is an equation involving a squared term, such as $c^2$ . In trig equations, you might make a quadratic in $sin⁡x\sin x$ or $cos⁡x\cos x$ .

Example

Changing to a quadratic in cosine

Solve $3\sin^2 x=5\cos x+1$ for $\le x < 360^\circ$ . Give your answers to one decimal place.

Since the right-hand side contains $cos⁡x\cos x$ , rewrite $sin^2 x$ using $sin^2 x=1-\cos^2 x$ :
$3(1-\cos^2 x)=5\cos x+1$
Rearrange into a quadratic:
$3−3cos⁡2x=5cos⁡x+13cos⁡2x+5cos⁡x−2=0\begin{aligned} 3-3\cos^2 x &= 5\cos x+1 \\ 3\cos^2 x+5\cos x-2 &= 0 \end{aligned}$
Factorise the quadratic:
$(3cos⁡x−1)(cos⁡x+2)=0(3\cos x-1)(\cos x+2)=0$
Solve each factor:
$cos⁡x=13orcos⁡x=−2\cos x=\frac{1}{3}\quad \text{or}\quad \cos x=-2$
Reject $cos⁡x=−2\cos x=-2$ because cosine values must be between -1 and 1. Now solve $cos⁡x=13\cos x=\frac{1}{3}$ :

$Cosine values must lie between -1 and 1, so \cos x=-2 is impossible for a real angle.$

$$
x=70.5287\ldots^\circ,\quad 289.4712\ldots^\circ
$$

6. Round to one decimal place:

$$
x=70.5^\circ,\ 289.5^\circ
$$

Common Mistake

Check possible trig values

For real angles, $sin⁡x\sin x$ and $cos⁡x\cos x$ must lie between -1 and 1. If a quadratic gives $sin⁡x=1.4\sin x=1.4$ or $cos⁡x=−2\cos x=-2$ , reject that branch.

Factorising without losing solutions

Some equations mix tangent with sine or cosine. The safest method is usually to rewrite $tan⁡x\tan x$ as $sin⁡xcos⁡x\frac{\sin x}{\cos x}$ , then factorise.

Example

Do not divide away a solution

Solve $2tan⁡x=3sin⁡x2\tan x=3\sin x$ for $\le x < 360^\circ$ . Give your answers to one decimal place where appropriate.

Rewrite tangent as sine over cosine:
$2sin⁡xcos⁡x=3sin⁡x\frac{2\sin x}{\cos x}=3\sin x$
Multiply by $cos⁡x\cos x$ . The original equation is already undefined when $cos⁡x=0\cos x=0$ , so 90° and 270° cannot be answers:
$2sin⁡x=3sin⁡xcos⁡x2\sin x=3\sin x\cos x$
Bring everything to one side and factorise:

$Factorising creates two branches of solutions, avoiding the loss of answers that would happen if dividing by \sin x.$

$$
\sin x(2-3\cos x)=0
$$

4. Solve each factor:

$$
\sin x=0\quad \text{or}\quad \cos x=\frac{2}{3}
$$

5. Find the values in $\le x < 360^\circ$ :

$$
x=0^\circ,\ 48.1896\ldots^\circ,\ 180^\circ,\ 311.8103\ldots^\circ
$$

6. Round where needed:

$$
x=0^\circ,\ 48.2^\circ,\ 180^\circ,\ 311.8^\circ
$$

Common Mistake

Dividing by a trig factor

If you divide both sides by $sin⁡x\sin x$ , you lose the solutions where $sin⁡x=0\sin x=0$ . Factorising keeps those solutions visible.

Double-angle equations

A double angle is an angle such as $2 x$ . Treat $2 x$ as the angle you are solving for first, then divide by 2 at the end.

Example

Solving a double-angle equation

Solve $2sin⁡(2x)tan⁡(2x)=cos⁡(2x)+12\sin(2x)\tan(2x)=\cos(2x)+1$ for $\le x < 180^\circ$ . Give your answers to two decimal places.

Let $c=cos⁡(2x)c=\cos(2x)$ . Also use $tan⁡(2x)=sin⁡(2x)cos⁡(2x)\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$ and $sin^2(2x)=1-\cos^2(2x)$ :
$2(1−c2)c=c+1\frac{2(1-c^2)}{c}=c+1$
Multiply by $c$ and rearrange:
$2−2c2=c2+c3c2+c−2=0\begin{aligned} 2-2c^2 &= c^2+c \\ 3c^2+c-2 &= 0 \end{aligned}$
Factorise:
$(3 c - 2) (c + 1) = 0$
So:
$cos⁡(2x)=23orcos⁡(2x)=−1\cos(2x)=\frac{2}{3}\quad \text{or}\quad \cos(2x)=-1$
Since $\le x < 180^\circ$ , the inside angle satisfies $\le 2x < 360^\circ$ :

For a double-angle equation, solve for 2x over the doubled interval before halving the answers.

$$
2x=48.1896\ldots^\circ,\ 180^\circ,\ 311.8103\ldots^\circ
$$

6. Divide by 2:

$$
x=24.09^\circ,\ 90^\circ,\ 155.91^\circ
$$

Exam technique

In the exam

Write down the interval for the inside angle before using your calculator.
Use symmetry or the period of the trig function to find every possible angle.
If the equation can be factorised, factorise instead of dividing by a trig expression.
Substitute or mentally check your answers in the original equation, especially if you multiplied by something involving trig.

Self review

Check yourself

Which trig functions have period 360°, and which has period 180°?
Why does $tan^2 x=4$ lead to two separate equations?
In an equation like $4sin⁡xcos⁡x=2sin⁡x4\sin x\cos x=2\sin x$ , why might dividing by $sin⁡x\sin x$ be dangerous?

10.4 Solving Trigonometric Equations

What you'll learn

How to find all solutions to trigonometric equations in a given interval.
How to handle shifted or stretched angles such as $2x+10∘2x+10^\circ$ .
How to solve equations involving $sin^2 x$ , $cos^2 x$ or $tan^2 x$ .
How to use identities and factorisation without losing valid answers.

The big idea: one trig equation can have several answers

A trigonometric equation is an equation involving a trig function such as $sin⁡x\sin x$ , $cos⁡x\cos x$ or $tan⁡x\tan x$ .

In this topic, angles are measured in degrees, so make sure your calculator is in degree mode. The key challenge is that trig graphs repeat, so one calculator answer is usually not enough.

Definition

Solution interval

A graph helps you see why there can be more than one answer. Solving $sin⁡(x−30∘)=0.3\sin(x-30^\circ)=0.3$ means finding where the sine curve meets the horizontal line $y = 0.3$ .

The shifted sine graph intersects the line y=0.3 more than once, showing why a trigonometric equation can have several solutions.

Graph of y = sin(x - 30°) and y = 0.3 showing two intersections

Key Idea

The calculator gives a starting point

Your calculator gives a principal value, which is the first angle it finds. You must then use symmetry and periodicity to find all the other angles in the required interval.

A reliable routine for shifted sine and cosine equations

For equations like $cos⁡(2x+10∘)=0.4\cos(2x+10^\circ)=0.4$ , the expression inside the trig function is not just $x$ . A good method is to temporarily replace the inside angle with a new letter.

Definition

Period

The period of a trig function is how long it takes before the graph repeats. Sine and cosine repeat every 360°, while tangent repeats every 180°.

Example

Solving a shifted cosine equation

Solve $cos⁡(2x+10∘)=0.4\cos(2x+10^\circ)=0.4$ for $\le x < 180^\circ$ . Give your answers to one decimal place.

Let the inside angle be $u$ :
$2x+10^\circ$
Convert the interval for $x$ into an interval for $u$ . If $\le x < 180^\circ$ , then:

$The substitution u=2x+10^\circ stretches and shifts the original x-interval to the corresponding u-interval.$

$$
10^\circ \le u < 370^\circ
$$

3. Use your calculator to find the principal value:

$$
\arccos(0.4)=66.4218\ldots^\circ
$$

4. Cosine is positive in quadrants I and IV, so within $10∘≤u<370∘10^\circ \le u < 370^\circ$ :

Cosine is positive in quadrants I and IV, giving two possible u-angles in the interval.

$$
u = 66.4218\ldots^\circ,\quad 293.5781\ldots^\circ
$$

5. Convert back to $x$ using $u=2x+10∘u=2x+10^\circ$ :

$$
\begin{aligned}
x &= \frac{66.4218\ldots-10}{2}=28.2109\ldots^\circ \\
x &= \frac{293.5781\ldots-10}{2}=141.7890\ldots^\circ
\end{aligned}
$$

6. Round to one decimal place:

$$
x = 28.2^\circ,\ 141.8^\circ
$$

Tip

Work with the inside angle first

If the equation contains $sin⁡(3θ−20∘)\sin(3\theta-20^\circ)$ , solve for the whole angle $3θ−20∘3\theta-20^\circ$ first. Only divide by 3 and add 20° at the end.

Tangent equations: remember the shorter period

Example

Solving a shifted tangent equation

Solve $tan⁡(θ+40∘)=−1.8\tan(\theta+40^\circ)=-1.8$ for $−180∘≤θ<180∘-180^\circ \le \theta < 180^\circ$ . Give your answers to one decimal place.

Let the inside angle be $u$ :
$u=θ+40∘u=\theta+40^\circ$
Convert the interval. If $−180∘≤θ<180∘-180^\circ \le \theta < 180^\circ$ , then:
$−140∘≤u<220∘-140^\circ \le u < 220^\circ$
Find one calculator solution:
$arctan⁡(−1.8)=−60.9453…∘\arctan(-1.8)=-60.9453\ldots^\circ$
Since tangent has period 180°, add or subtract 180° until you have all values in the interval:

$Tangent solutions repeat every 180^\circ, so the calculator solution must be shifted by multiples of 180^\circ within the allowed interval.$

$$
u=-60.9453\ldots^\circ,\quad 119.0546\ldots^\circ
$$

5. Convert back to $θ\theta$ using $θ=u−40∘\theta=u-40^\circ$ :

$$
\theta=-100.9453\ldots^\circ,\quad 79.0546\ldots^\circ
$$

6. Round to one decimal place:

$$
\theta=-100.9^\circ,\ 79.1^\circ
$$

Squared trig functions

A squared trig function means the whole trig value is squared. For example, $tan^2 x$ means $tan x)^2$ , not $tan(x^2)$ .

When you square-root both sides, remember the positive and negative possibilities.

Example

Solving a squared tangent equation

Solve $tan^2 x=5$ for $\le x < 360^\circ$ . Give your answers to one decimal place.

Square-root both sides:
$tan⁡x=±5\tan x=\pm\sqrt{5}$
Find the reference angle:

The four tangent solutions come from using the same reference angle in all four quadrants after taking both positive and negative square roots.

$$
\arctan(\sqrt{5})=65.9051\ldots^\circ
$$

3. For $tan⁡x=5\tan x=\sqrt{5}$ , tangent is positive in quadrants I and III:

$$
x=65.9051\ldots^\circ,\quad 245.9051\ldots^\circ
$$

4. For $tan⁡x=−5\tan x=-\sqrt{5}$ , tangent is negative in quadrants II and IV:

$$
x=114.0948\ldots^\circ,\quad 294.0948\ldots^\circ
$$

5. Round to one decimal place:

$$
x=65.9^\circ,\ 114.1^\circ,\ 245.9^\circ,\ 294.1^\circ
$$

Common Mistake

Forgetting the negative square root

From $tan^2 x=5$ , do not only write $tan⁡x=5\tan x=\sqrt{5}$ . You also need $tan⁡x=−5\tan x=-\sqrt{5}$ .

Using identities to make quadratics

An identity is an equation that is true for all allowed values of the variable. The most useful identity here is:

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

So:

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

and:

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

A quadratic equation is an equation involving a squared term, such as $c^2$ . In trig equations, you might make a quadratic in $sin⁡x\sin x$ or $cos⁡x\cos x$ .

Example

Changing to a quadratic in cosine

Solve $3\sin^2 x=5\cos x+1$ for $\le x < 360^\circ$ . Give your answers to one decimal place.

Since the right-hand side contains $cos⁡x\cos x$ , rewrite $sin^2 x$ using $sin^2 x=1-\cos^2 x$ :
$3(1-\cos^2 x)=5\cos x+1$
Rearrange into a quadratic:
$3−3cos⁡2x=5cos⁡x+13cos⁡2x+5cos⁡x−2=0\begin{aligned} 3-3\cos^2 x &= 5\cos x+1 \\ 3\cos^2 x+5\cos x-2 &= 0 \end{aligned}$
Factorise the quadratic:
$(3cos⁡x−1)(cos⁡x+2)=0(3\cos x-1)(\cos x+2)=0$
Solve each factor:
$cos⁡x=13orcos⁡x=−2\cos x=\frac{1}{3}\quad \text{or}\quad \cos x=-2$
Reject $cos⁡x=−2\cos x=-2$ because cosine values must be between -1 and 1. Now solve $cos⁡x=13\cos x=\frac{1}{3}$ :

$Cosine values must lie between -1 and 1, so \cos x=-2 is impossible for a real angle.$

$$
x=70.5287\ldots^\circ,\quad 289.4712\ldots^\circ
$$

6. Round to one decimal place:

$$
x=70.5^\circ,\ 289.5^\circ
$$

Common Mistake

Check possible trig values

For real angles, $sin⁡x\sin x$ and $cos⁡x\cos x$ must lie between -1 and 1. If a quadratic gives $sin⁡x=1.4\sin x=1.4$ or $cos⁡x=−2\cos x=-2$ , reject that branch.

Factorising without losing solutions

Some equations mix tangent with sine or cosine. The safest method is usually to rewrite $tan⁡x\tan x$ as $sin⁡xcos⁡x\frac{\sin x}{\cos x}$ , then factorise.

Example

Do not divide away a solution

Solve $2tan⁡x=3sin⁡x2\tan x=3\sin x$ for $\le x < 360^\circ$ . Give your answers to one decimal place where appropriate.

Rewrite tangent as sine over cosine:
$2sin⁡xcos⁡x=3sin⁡x\frac{2\sin x}{\cos x}=3\sin x$
Multiply by $cos⁡x\cos x$ . The original equation is already undefined when $cos⁡x=0\cos x=0$ , so 90° and 270° cannot be answers:
$2sin⁡x=3sin⁡xcos⁡x2\sin x=3\sin x\cos x$
Bring everything to one side and factorise:

$Factorising creates two branches of solutions, avoiding the loss of answers that would happen if dividing by \sin x.$

$$
\sin x(2-3\cos x)=0
$$

4. Solve each factor:

$$
\sin x=0\quad \text{or}\quad \cos x=\frac{2}{3}
$$

5. Find the values in $\le x < 360^\circ$ :

$$
x=0^\circ,\ 48.1896\ldots^\circ,\ 180^\circ,\ 311.8103\ldots^\circ
$$

6. Round where needed:

$$
x=0^\circ,\ 48.2^\circ,\ 180^\circ,\ 311.8^\circ
$$

Common Mistake

Dividing by a trig factor

If you divide both sides by $sin⁡x\sin x$ , you lose the solutions where $sin⁡x=0\sin x=0$ . Factorising keeps those solutions visible.

Double-angle equations

A double angle is an angle such as $2 x$ . Treat $2 x$ as the angle you are solving for first, then divide by 2 at the end.

Example

Solving a double-angle equation

Solve $2sin⁡(2x)tan⁡(2x)=cos⁡(2x)+12\sin(2x)\tan(2x)=\cos(2x)+1$ for $\le x < 180^\circ$ . Give your answers to two decimal places.

Let $c=cos⁡(2x)c=\cos(2x)$ . Also use $tan⁡(2x)=sin⁡(2x)cos⁡(2x)\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$ and $sin^2(2x)=1-\cos^2(2x)$ :
$2(1−c2)c=c+1\frac{2(1-c^2)}{c}=c+1$
Multiply by $c$ and rearrange:
$2−2c2=c2+c3c2+c−2=0\begin{aligned} 2-2c^2 &= c^2+c \\ 3c^2+c-2 &= 0 \end{aligned}$
Factorise:
$(3 c - 2) (c + 1) = 0$
So:
$cos⁡(2x)=23orcos⁡(2x)=−1\cos(2x)=\frac{2}{3}\quad \text{or}\quad \cos(2x)=-1$
Since $\le x < 180^\circ$ , the inside angle satisfies $\le 2x < 360^\circ$ :

For a double-angle equation, solve for 2x over the doubled interval before halving the answers.

$$
2x=48.1896\ldots^\circ,\ 180^\circ,\ 311.8103\ldots^\circ
$$

6. Divide by 2:

$$
x=24.09^\circ,\ 90^\circ,\ 155.91^\circ
$$

Exam technique

In the exam

Write down the interval for the inside angle before using your calculator.
Use symmetry or the period of the trig function to find every possible angle.
If the equation can be factorised, factorise instead of dividing by a trig expression.
Substitute or mentally check your answers in the original equation, especially if you multiplied by something involving trig.

Self review

Check yourself

Which trig functions have period 360°, and which has period 180°?
Why does $tan^2 x=4$ lead to two separate equations?
In an equation like $4sin⁡xcos⁡x=2sin⁡x4\sin x\cos x=2\sin x$ , why might dividing by $sin⁡x\sin x$ be dangerous?

10.4 Solving Trigonometric Equations

What you'll learn

The big idea: one trig equation can have several answers

A reliable routine for shifted sine and cosine equations

Tangent equations: remember the shorter period

Squared trig functions

Using identities to make quadratics

Factorising without losing solutions

Double-angle equations

Recap questions

You've reached the end

10.4 Solving Trigonometric Equations

What you'll learn

The big idea: one trig equation can have several answers

A reliable routine for shifted sine and cosine equations

Tangent equations: remember the shorter period

Squared trig functions

Using identities to make quadratics

Factorising without losing solutions

Double-angle equations

Recap questions

You've reached the end

10.4 Solving Trigonometric Equations

What you'll learn

The big idea: one trig equation can have several answers

A reliable routine for shifted sine and cosine equations

Tangent equations: remember the shorter period

Squared trig functions

Using identities to make quadratics

Factorising without losing solutions

Double-angle equations

Recap questions

You've reached the end

10.4 Solving Trigonometric Equations Revision Guide

10.4 Solving Trigonometric Equations

What you'll learn

The big idea: one trig equation can have several answers

A reliable routine for shifted sine and cosine equations

Tangent equations: remember the shorter period

Squared trig functions

Using identities to make quadratics

Factorising without losing solutions

Double-angle equations

Recap questions

You've reached the end

10.4 Solving Trigonometric Equations Revision Guide