Space, shape and measurement: Solve problems by constructing and interpreting trigonometric models

# Unit 2: Solve trigonometric equations

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Solve equations involving double and compound angles.

## What you should know

Before you start this unit, make sure you can:

• State and apply the reduction formulae for $\scriptsize {{180}^\circ}\pm \theta$, $\scriptsize {{90}^\circ}\pm \theta$ and $\scriptsize {{360}^\circ}\pm \theta$. Refer to level 3 subject outcome 3.3 unit 2 if you need help with this.
• State and apply the basic trigonometric identities of $\scriptsize \tan \theta =\displaystyle \frac{{\sin \theta }}{{\cos \theta }}$ and $\scriptsize {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Refer to level 3 subject outcome 3.3 unit 3 if you need help with this.
• State and apply the compound and double angle identities. Refer to unit 1 of this subject outcome if you need help with this.
• Find the general solutions of trigonometric equations. Refer to level 3 subject outcome 3.3 unit 4 if you need help with this.

## Introduction

In level 3 subject outcome 3.3 unit 4, we learnt how to find the general solution of an equation that involved a trigonometric ratio. If you need to, you should review this unit before continuing.

Remember that the trigonometric functions are repetitive. Sine and cosine repeat every $\scriptsize {{360}^\circ}$ (once a full revolution) and tangent repeats every $\scriptsize {{180}^\circ}$ (twice every revolution). We say that the trigonometric functions are periodic. Sine and cosine functions have periods of $\scriptsize {{360}^\circ}$ and the tangent function has a period of $\scriptsize {{180}^\circ}$.

## The general solution revision

Because the trigonometric functions are periodic (they repeat themselves), when we solve an equation such as $\scriptsize \sin \theta =\displaystyle \frac{1}{2}$, there is not only one solution. We may know that $\scriptsize \sin {{30}^\circ}=\displaystyle \frac{1}{2}$ but, for example, so does $\scriptsize \sin ({{30}^\circ}+{{360}^\circ})=\sin {{390}^\circ}$, $\scriptsize \sin ({{30}^\circ}+2\times {{360}^\circ})=\sin {{750}^\circ}$ and $\scriptsize \sin ({{30}^\circ}-{{360}^\circ})=\sin -{{330}^\circ}$.

We also know that $\scriptsize \sin {{150}^\circ}=\displaystyle \frac{1}{2}$ but so do $\scriptsize \sin ({{150}^\circ}+{{360}^\circ})=\sin {{510}^\circ}$, $\scriptsize \sin ({{150}^\circ}+2\times {{360}^\circ})=\sin {{870}^\circ}$ and $\scriptsize \sin ({{150}^\circ}-{{360}^\circ})=\sin -{{210}^\circ}$, for example.

The general solution to the equation $\scriptsize \sin \theta =\displaystyle \frac{1}{2}$ is $\scriptsize \theta ={{30}^\circ}+k\cdot {{360}^\circ}\text{ or }\theta ={{150}^\circ}+k\cdot {{360}^\circ},k\in \mathbb{Z}$. In other words, $\scriptsize \theta$ is equal to $\scriptsize {{30}^\circ}$ plus or minus any integer multiple of $\scriptsize {{360}^\circ}$ or $\scriptsize \theta$ is equal to $\scriptsize {{150}^\circ}$ plus or minus any integer multiple of $\scriptsize {{360}^\circ}$.

When we learnt how to find the general solution of trigonometric equations, we called the answer we get for $\scriptsize \theta$ from the calculator (e.g. $\scriptsize \theta ={{30}^\circ}$) the reference angle. It is the basis upon which we build the general solution.

Work through the next two examples and the exercise that follows to remind yourself how to find the general solution of simple trigonometric equations.

### Example 2.1

Determine the general solution for $\scriptsize \sin \theta =0.6$.

Solution

Step 1: Use a calculator to determine the reference angle
\scriptsize \begin{align*}\sin \theta &=0.6\\\therefore \theta &={{36.9}^\circ}\end{align*}
Note: Unless told otherwise, we usually round the reference angle to one decimal place.

Step 2: Use the CAST diagram to determine any other possible solutions
$\scriptsize \sin \theta =0.6$. In other words, sine is positive. Sine is positive in the first and second quadrants. We already have the first quadrant solution (the reference angle of $\scriptsize \theta ={{36.9}^\circ}$). We need to find the second quadrant solution. We know that $\scriptsize \sin ({{180}^\circ}-\theta )=\sin \theta$. Therefore, the second quadrant solution is $\scriptsize {{180}^\circ}-{{36.9}^\circ}={{143.1}^\circ}$.

Step 3: Generate the general solution
$\scriptsize \theta ={{36.9}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{143.1}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }$

Step 4: Check your general solution
It is always a good idea to check that your final solutions satisfy the original equation. Choose a random value for $\scriptsize k$.

$\scriptsize k=-2$:
\scriptsize \begin{align*}\theta &={{36.9}^\circ}-2\times {{360}^\circ}\text{ or }\theta ={{143.1}^\circ}-2\times {{360}^\circ}\\\therefore \theta &=-{{683.1}^\circ}\text{ or }\theta =-{{576.9}^\circ}\end{align*}
$\scriptsize \sin (-{{683.1}^\circ})=0.6$
$\scriptsize \sin (-{{576.9}^\circ})=0.6$

Our general solution is correct.

### Example 2.2

Determine the general solution for $\scriptsize 7\cos 2\theta +4=0$.

Solution

\scriptsize \displaystyle \begin{align*}7\cos 2\theta +4&=0\\\therefore \cos 2\theta &=-\displaystyle \frac{4}{7}\end{align*}

Step 1: Use a calculator to determine the reference angle
\scriptsize \displaystyle \begin{align*}\cos 2\theta &=-\displaystyle \frac{4}{7}\\\therefore 2\theta &={{124.8}^\circ}\end{align*}

Note: We keep working with the reference angle of $\scriptsize 2\theta$ until we generate the general solution.

Step 2: Use the CAST diagram to determine any other possible solutions
Our equation is $\scriptsize \displaystyle \cos 2\theta =-\displaystyle \frac{4}{7}$. $\scriptsize \cos 2\theta \lt 0$. Cosine is negative in the second and third quadrants. Our reference angle is in the second quadrant.
Second quadrant: $\scriptsize 2\theta ={{124.8}^\circ}$

Third quadrant: $\scriptsize \cos ({{360}^\circ}-\theta )=\cos \theta$
$\scriptsize 2\theta ={{360}^\circ}-{{124.8}^\circ}={{235.2}^\circ}$.

Step 3: Generate the general solution
\scriptsize \begin{align*}2\theta & ={{124.8}^\circ}+k{{.360}^\circ}\text{ or 2}\theta ={{235.2}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }\\\therefore \theta & ={{62.4}^\circ}+k{{.180}^\circ}\text{ or }\theta ={{117.6}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }\end{align*}

Step 4: Check your general solution
$\scriptsize k=2$:
$\scriptsize \therefore \theta ={{62.4}^\circ}+2\times {{180}^\circ}={{422.4}^\circ}\text{ or }\theta ={{117.6}^\circ}+2\times {{180}^\circ}={{477.6}^\circ}$
$\scriptsize \displaystyle \cos (2\times {{422.4}^\circ})=-\displaystyle \frac{4}{7}$
$\scriptsize \displaystyle \cos (2\times {{477.6}^\circ})=-\displaystyle \frac{4}{7}$

### Take note!

The general solutions for equations involving the three basic trigonometric ratios can be written as follows:

If $\scriptsize \sin \theta =x$ then:
$\scriptsize \theta =\left( {{{{\sin }}^{{-1}}}x+k{{{.360}}^\circ}} \right)\text{ or }\theta =\left( {\left( {{{{180}}^\circ}-{{{\sin }}^{{-1}}}x} \right)+k{{{.360}}^\circ}} \right),k\in \mathbb{Z}$

If $\scriptsize \cos \theta =x$ then:
$\scriptsize \theta =\left( {{{{\cos }}^{{-1}}}x+k{{{.360}}^\circ}} \right)\text{ or }\theta =\left( {\left( {{{{360}}^\circ}-{{{\cos }}^{{-1}}}x} \right)+k{{{.360}}^\circ}} \right),k\in \mathbb{Z}$

If $\scriptsize \tan \theta =x$ then:
$\scriptsize \theta =\left( {{{{\tan }}^{{-1}}}x+k{{{.180}}^\circ}} \right),k\in \mathbb{Z}$

### Exercise 2.1

1. Find the general solution for $\scriptsize \theta$ in the following equations:
1. $\scriptsize \cos \theta =0.45$
2. $\scriptsize 2\tan \theta =-5$
3. $\scriptsize 8\sin \theta +3=0$
4. $\scriptsize -6\cos 2\theta -3=0$
2. Determine $\scriptsize \theta$ for the given interval:
1. $\scriptsize 3\cos \theta -2=-3$ for the interval $\scriptsize [{{0}^\circ},{{360}^\circ}]$
2. $\scriptsize \tan (3\theta -{{42}^\circ})=3.4$ where $\scriptsize \theta \in [{{0}^\circ},{{180}^\circ}]$

The full solutions are at the end of the unit.

## Trigonometric equations with double and compound angles

We can use the compound and double angle identities we learnt about in unit 1 to help us solve some more complicated trigonometric equations. They may look complicated to begin with, but once we have applied the compound and double angle identities, they become much simpler.

### Example 2.3

Determine the general solution for $\scriptsize \theta$ in $\scriptsize \displaystyle \frac{{1-\sin \theta -\cos 2\theta }}{{\sin 2\theta -\cos \theta }}=-1$.

Solution

Before we start solving any trigonometric equation, we need to try and simplify it as far as possible. In this case, there are double angles involved. Therefore, we can use the double angle identities. So we start with the LHS of the equation and simplify it.
\scriptsize \begin{align*}\displaystyle \frac{{1-\sin \theta -\cos 2\theta }}{{\sin 2\theta -\cos \theta }}&=\displaystyle \frac{{1-\sin \theta -\left( {1-2{{{\sin }}^{2}}\theta } \right)}}{{2\sin \theta \cos \theta -\cos \theta }}\\&=\displaystyle \frac{{1-\sin \theta -1+2{{{\sin }}^{2}}\theta }}{{2\sin \theta \cos \theta -\cos \theta }}\\&=\displaystyle \frac{{2{{{\sin }}^{2}}\theta -\sin \theta }}{{2\sin \theta \cos \theta -\cos \theta }}\\&=\displaystyle \frac{{\sin \theta (2\sin \theta -1)}}{{\cos \theta (2\sin \theta -1)}}\\&=\displaystyle \frac{{\sin \theta }}{{\cos \theta }}\\&=\tan \theta \end{align*}
Now our equation becomes easy to solve: $\scriptsize \tan \theta =-1$.

Ref angle: $\scriptsize \theta =-{{45}^\circ}$

General solution: $\scriptsize \theta =-{{45}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$

Note: We can also write the general solution with a positive angle as $\scriptsize \theta ={{135}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$.

### Example 2.4

Prove that $\scriptsize 4\sin \theta {{\cos }^{3}}\theta -4{{\sin }^{3}}\theta \cos \theta =\sin 4\theta$ and hence determine the general solution for $\scriptsize \theta$ in $\scriptsize 4\sin \theta {{\cos }^{3}}\theta -4{{\sin }^{3}}\theta \cos \theta =0.8$.

Solution

In this example, we are first asked to prove that $\scriptsize 4\sin \theta {{\cos }^{3}}\theta -4{{\sin }^{3}}\theta \cos \theta =\sin 4\theta$.
\scriptsize \begin{align*}\text{LHS}&=4\sin \theta {{\cos }^{3}}\theta -4{{\sin }^{3}}\theta \cos \theta \\&=4\sin \theta \cos \theta ({{\cos }^{2}}\theta -{{\sin }^{2}}\theta )\\&=2\times 2\sin \theta \cos \theta \times \cos 2\theta \\&=2\times \sin 2\theta \times \cos 2\theta \\&=\sin 4\theta \end{align*}
Now we can solve the equation:
\scriptsize \displaystyle \begin{align*}4\sin \theta {{\cos }^{3}}\theta -4{{\sin }^{3}}\theta \cos \theta & =0.8\\\therefore \sin 4\theta & =0.8\end{align*}
Ref angle: $\scriptsize 4\theta ={{53.1}^\circ}$

Sine is positive in the first and second quadrants.
$\scriptsize 4\theta ={{180}^\circ}-{{53.1}^\circ}={{126.9}^\circ}$

General solution:
\scriptsize \begin{align*}4\theta & ={{53.1}^\circ}+k{{.360}^\circ}\text{ or }4\theta ={{126.9}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\\\therefore \theta & ={{13.275}^\circ}+k{{.90}^\circ}\text{ or }\theta ={{31.725}^\circ}+k{{.90}^\circ},k\in \mathbb{Z}\end{align*}

### Example 2.5

Find the general solution for $\scriptsize {{\sin }^{2}}\theta \cos \theta ={{\cos }^{3}}\theta$.

Solution

We need to be careful here. We cannot divide through by $\scriptsize \sin \theta$ or $\scriptsize {{\sin }^{2}}\theta$. If we do so, we will lose some of the solution. Instead, we need to proceed as if we were solving a quadratic equation.
\scriptsize \begin{align*}{{\sin }^{2}}\theta \cos \theta &={{\cos }^{3}}\theta \\\therefore {{\sin }^{2}}\theta \cos \theta -{{\cos }^{3}}\theta =0\\\therefore \cos \theta ({{\sin }^{2}}\theta -{{\cos }^{2}}\theta )&=0\quad \text{Take a factor of }-1\text{ out of the bracket}\\\therefore -\cos \theta ({{\cos }^{2}}\theta -{{\sin }^{2}}\theta )&=0\\\therefore \cos \theta ({{\cos }^{2}}\theta -{{\sin }^{2}}\theta )&=0\\\therefore \cos \theta \times \cos 2\theta &=0\\\therefore \cos \theta &=0\text{ or }\cos 2\theta =0\end{align*}

We can deal with each part of the solution separately.

$\scriptsize \cos \theta =0$:
Ref angle: $\scriptsize \theta ={{90}^\circ}$

General solution: $\scriptsize \theta ={{90}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{270}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$.

We can simplify this general solution to $\scriptsize \theta ={{90}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$.

$\scriptsize \cos 2\theta =0$:
Ref angle: $\scriptsize 2\theta ={{90}^\circ}$

General solution:
\scriptsize \begin{align*}2\theta & ={{90}^\circ}+k{{.360}^\circ}\text{ or 2}\theta ={{270}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\\\therefore \theta & ={{45}^\circ}+k{{.180}^\circ}\text{ or }\theta ={{135}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\end{align*}.

Again, we can simplify this general solution to $\scriptsize \theta ={{45}^\circ}+k{{.90}^\circ},k\in \mathbb{Z}$.

### Take note!

There are two basic strategies for solving these more complicated trigonometric equations:

1. Simplify one side of the equation down to a single trigonometric ratio using the various trigonometric identities at your disposal.
2. Make the one side of the equation equal to zero and then simplify the other side in order to make use of the zero product rule – if $\scriptsize a\times b=0$ then either $\scriptsize a=0$ or $\scriptsize b=0$. This is the technique we use to solve quadratic equations.

### Exercise 2.2

1. Determine the general solution in each case:
1. $\scriptsize \cos 2x=\sin {{32}^\circ}$
2. $\scriptsize \cos 2x=\sin x$
3. $\scriptsize \sin \theta \cdot \sin 2\theta +\cos 2\theta =1$
4. $\scriptsize 5\tan 2x-1={{\tan }^{2}}2x+5$

Question 2 adapted from Everything Maths Grade 12 Exercise 4-4 question 5a

1. Given$\scriptsize \sin x\cos 3x-\cos x\sin 3x=\tan {{140}^\circ}$:
1. Find the general solution.
2. Determine the solutions for the interval $\scriptsize \left[ {{{0}^\circ},{{{90}}^\circ}} \right]$.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to apply the trigonometric identities, especially the compound and double angle identities, to solve more complicated trigonometric equations.

# Unit 2: Assessment

#### Suggested time to complete: 45 minutes

1. Solve each equation for each given interval. If no interval is given, find the general solution.
1. $\scriptsize 2\cos \theta -1=0$
2. $\scriptsize {{\sin }^{2}}\theta +\sin \theta -1=0$
3. $\scriptsize 4\sin x+3\tan x=\displaystyle \frac{3}{{\cos x}}+4$ for $\scriptsize \left[ {{{0}^\circ},{{{180}}^\circ}} \right]$
4. $\scriptsize 6\cos \theta -5=\displaystyle \frac{4}{{\cos \theta }}$
5. $\scriptsize \cos 2x-\cos x+1=0$ for $\scriptsize \left[ {{{0}^\circ},{{{360}}^\circ}} \right]$

Question 2 adapted from NC(V) Mathematics Level 4 Paper 2November 2015 question 2.5

1. Given that $\scriptsize \cos 2\theta =2{{\cos }^{2}}\theta -1$:
1. Show that $\scriptsize \cos 2\theta +3\cos \theta -1=2{{\cos }^{2}}\theta +3\cos \theta -2$.
2. Hence, determine the value(s) of $\scriptsize \theta$ if $\scriptsize \cos 2\theta +3\cos \theta -1=0$ and $\scriptsize {{0}^\circ}\le \theta \le {{360}^\circ}$.

Question 3 adapted from NC(V) Mathematics Level 4 Paper 2 November 2016 question 2.5

1. If $\scriptsize 3\cos 2x-\sin 2x-1=0$ find the value(s) of $\scriptsize x$ in the interval $\scriptsize \left[ {{{0}^\circ},{{{360}}^\circ}} \right]$.

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

1. .
1. $\scriptsize \cos \theta =0.45$
Ref angle: $\scriptsize \theta ={{63.3}^\circ}$
$\scriptsize \cos \theta$ is positive in the first and fourth quadrants.
$\scriptsize \theta ={{360}^\circ}-{{63.3}^\circ}={{296.7}^\circ}$
General solution: $\scriptsize \theta ={{63.3}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{296.7}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }$
2. .
\scriptsize \begin{align*}2\tan \theta & =-5\\\therefore \tan \theta & =-\displaystyle \frac{5}{2}\end{align*}
Ref angle: $\scriptsize \theta =-{{68.2}^\circ}$
Note: You can also express your reference angle as a positive angle – $\scriptsize {{360}^\circ}-{{68.2}^\circ}={{291.8}^\circ}$
General solution: $\scriptsize \theta =-{{68.2}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }$
3. .
\scriptsize \displaystyle \begin{align*}8\sin \theta +3 & =0\\\therefore \sin \theta & =-\displaystyle \frac{3}{8}\end{align*}
Ref angle: $\scriptsize \theta =-{{22.0}^\circ}$
Note: You can also express your reference angle as a positive angle – $\scriptsize {{360}^\circ}-{{22.0}^\circ}={{338}^\circ}$
$\scriptsize \sin \theta$ is negative in the third and fourth quadrants.
$\scriptsize \theta ={{180}^\circ}+{{22.0}^\circ}={{202.0}^\circ}$
General solution: $\scriptsize \theta =-{{22.0}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{202.0}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }$
4. .
\scriptsize \begin{align*}-6\cos 2\theta -3 & =0\\\therefore \cos 2\theta & =-0.5\end{align*}
Ref angle: $\scriptsize \theta ={{120}^\circ}$
$\scriptsize \cos \theta$ is negative in the second and third quadrants.
$\scriptsize \theta ={{360}^\circ}-{{120.0}^\circ}={{240}^\circ}$
General solution:
\scriptsize \begin{align*}2\theta & ={{120}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{240}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }\\\therefore \theta & ={{60}^\circ}+k{{.180}^\circ}\text{ or }\theta ={{120}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\end{align*}
2. .
1. .
\scriptsize \begin{align*}3\cos \theta -2 & =-3\\\therefore \cos \theta & =-\displaystyle \frac{1}{3}\end{align*}
Ref angle: $\scriptsize \theta ={{109.5}^\circ}$
$\scriptsize \cos \theta$ is negative in the second and third quadrants.
$\scriptsize \theta ={{360}^\circ}-{{109.5}^\circ}={{250.5}^\circ}$
General solution: $\scriptsize \theta ={{109.5}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{250.5}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }$
Specific solution: $\scriptsize \theta ={{109.5}^\circ}\text{ or }\theta ={{250.5}^\circ}$
2. $\scriptsize \tan (3\theta -{{42}^\circ})=3.4$
Ref angle: $\scriptsize 3\theta -{{42}^\circ}={{73.6}^\circ}$
General solution:
\scriptsize \begin{align*}3\theta -{{42}^\circ} &={{73.6}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\\\therefore 3\theta &={{115.6}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\\\therefore \theta & ={{38.5}^\circ}+k{{.60}^\circ},k\in \mathbb{Z}\end{align*}
Specific solution: $\scriptsize \theta ={{38.5}^\circ}\text{ or }\theta ={{98.5}^\circ}\text{ or }\theta ={{158.5}^\circ}$

Back to Exercise 2.1

### Exercise 2.2

1. .
1. .
\scriptsize \begin{align*}\cos 2x & =\sin {{32}^\circ}\\\therefore \cos 2x & =0.530\end{align*}
Or
\scriptsize \begin{align*}\cos 2x & =\sin {{32}^\circ}\\\therefore \cos 2x & =\cos ({{90}^\circ}-{{32}^\circ})\\&=\cos {{58}^\circ}\end{align*}
Ref angle: $\scriptsize \theta ={{58}^\circ}$
Cosine is positive in the first and fourth quadrants.
$\scriptsize \theta ={{360}^\circ}-{{58}^\circ}={{302}^\circ}$
General solution: $\scriptsize \theta ={{58}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{302}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
So:
\scriptsize \begin{align*}2x&={{58}^\circ}+k{{.360}^\circ}\text{ or }2x={{302}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\\x&={{29}^\circ}+k{{.180}^\circ}\text{ or }x={{151}^\circ}+k{{.180}^\circ}\text{, }k\in \mathbb{Z}\end{align*}
2. .
\scriptsize \begin{align*}\cos 2x & =\sin x\\\therefore 1-2{{\sin }^{2}}x & =\sin x\\\therefore 2{{\sin }^{2}}x+\sin x-1 & =0\\\therefore (2\sin x-1)(\sin x+1) & =0\\\therefore \sin x=\displaystyle \frac{1}{2}\text{ } & \text{or }\sin x=-1\end{align*}
$\scriptsize \sin x=\displaystyle \frac{1}{2}$:
Ref angle: $\scriptsize x={{30}^\circ}$
Sine is positive in the first and second quadrants.
$\scriptsize x={{180}^\circ}-{{30}^\circ}={{150}^\circ}$
General solution: $\scriptsize x={{30}^\circ}+k{{.360}^\circ}\text{ or }x={{150}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
$\scriptsize \sin x=-1$:
Ref angle: $\scriptsize x=-{{90}^\circ}$
General solution: $\scriptsize x={{270}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
3. .
\scriptsize \begin{align*}\sin \theta \cdot \sin 2\theta +\cos 2\theta & =1\\\therefore \sin \theta \cdot 2\sin \theta \cos \theta +{{\cos }^{2}}\theta -{{\sin }^{2}}\theta -1 & =0\\\therefore 2{{\sin }^{2}}\theta \cos \theta +{{\cos }^{2}}\theta -1-{{\sin }^{2}}\theta & =0\\\therefore 2{{\sin }^{2}}\theta \cos \theta -{{\sin }^{2}}\theta -{{\sin }^{2}}\theta & =0\\\therefore 2{{\sin }^{2}}\theta \cos \theta -2{{\sin }^{2}}\theta & =0\\\therefore 2{{\sin }^{2}}\theta (\cos \theta -1) & =0\\\therefore 2{{\sin }^{2}}\theta &=0\text{ } & \text{or cos}\theta =1\\\therefore \sin \theta &=0\text{ } & \text{or }\cos \theta =1\end{align*}
$\scriptsize \sin \theta =0$:
Ref angle: $\scriptsize \theta ={{0}^\circ}$
General solution: $\scriptsize \theta ={{0}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$
$\scriptsize \cos \theta =1$:
Ref angle: $\scriptsize \theta ={{0}^\circ}$
General solution: $\scriptsize \theta ={{0}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
Overall general solution: $\scriptsize \theta ={{0}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$
4. .
\scriptsize \begin{align*}5\tan 2x-1 & ={{\tan }^{2}}2x+5\\\therefore {{\tan }^{2}}2x-5\tan 2x+6 & =0\\\therefore (\tan 2x-3)(\tan 2x-2) & =0\\\therefore \tan 2x=3\text{ } & \text{or }\tan 2x=2\end{align*}
$\scriptsize \tan 2x=3$:
Ref angle: $\scriptsize 2x={{71.6}^\circ}$
General solution:
\scriptsize \begin{align*}2x & ={{71.57}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\\\therefore x & ={{35.8}^\circ}+k{{.90}^\circ},k\in \mathbb{Z}\end{align*}
$\scriptsize \tan 2x=2$:
Ref angle: $\scriptsize 2x={{63.4}^\circ}$
General solution:
\scriptsize \begin{align*}2x & ={{63.4}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\\\therefore x & ={{31.71}^\circ}+k{{.90}^\circ},k\in \mathbb{Z}\end{align*}
2. .
1. .
\scriptsize \begin{align*}\text{LHS}&=\sin x\cos 3x-\cos x\sin 3x\\&=\sin x\cos (2x+x)-\cos x\sin (2x+x)\\&=\sin x\left( {\cos 2x\cos x-\sin 2x\sin x} \right)-\cos x\left( {\sin 2x\cos x+\cos 2x\sin x} \right)\\&=\sin x\left( {\left( {2{{{\cos }}^{2}}x-1} \right)\cos x-2\sin x\cos x\sin x} \right)-\cos x\left( {2\sin x\cos x\cos x+(2{{{\cos }}^{2}}x-1)\sin x} \right)\\&=\sin x\left( {2{{{\cos }}^{3}}x-\cos x-2{{{\sin }}^{2}}x\cos x} \right)-\cos x\left( {2\sin x{{{\cos }}^{2}}x+2\sin x{{{\cos }}^{2}}x-\sin x} \right)\\&=2\sin x{{\cos }^{3}}x-\sin x\cos x-2{{\sin }^{3}}x\cos x-2\sin x{{\cos }^{3}}x-2\sin x{{\cos }^{3}}x+\sin x\cos x\\&=-2\sin x{{\cos }^{3}}x-2{{\sin }^{3}}x\cos x\\&=-2\sin x\cos x({{\cos }^{2}}x+{{\sin }^{2}}x)\\&=-\sin 2x\end{align*}
Therefore:
\scriptsize \begin{align*}-\sin 2x & =\tan {{140}^\circ}\\\therefore \sin 2x & =-\tan {{140}^\circ}\\ & =0.839\end{align*}
Ref angle: $\scriptsize 2x={{57.05}^\circ}$
Sine is positive in the first and second quadrants.
$\scriptsize 2x={{180}^\circ}-{{57.05}^\circ}={{122.95}^\circ}$
General solution:
\scriptsize \begin{align*}2x & =57.05+k{{.360}^\circ}\text{ or 2}x={{122.95}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\\\therefore x & ={{28.525}^\circ}+k{{.180}^\circ}\text{ or }x={{61.475}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\end{align*}
2. Solutions for the interval $\scriptsize \left[ {{{0}^\circ},{{{90}}^\circ}} \right]$: $\scriptsize x ={{28.525}^\circ}\text{ or }x={{61.475}^\circ}$

Back to Exercise 2.2

### Unit 2: Assessment

1. .
1. .
\scriptsize \begin{align*}2\cos \theta -1 & =0\\\therefore \cos \theta & =\displaystyle \frac{1}{2}\end{align*}
Ref angle: $\scriptsize \theta ={{60}^\circ}$
Cosine is positive in the first and fourth quadrants.
$\scriptsize \theta ={{360}^\circ}-{{60}^\circ}={{300}^\circ}$
General solution: $\scriptsize \theta ={{60}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{300}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
2. .
\scriptsize \begin{align*}2{{\sin }^{2}}\theta +\sin \theta -1 & =0\\\therefore (2\sin \theta -1)(\sin \theta +1) & =0\\\therefore \sin \theta =\displaystyle \frac{1}{2}\text{ } & \text{or }\sin \theta =-1\end{align*}
$\scriptsize \sin \theta =\displaystyle \frac{1}{2}$:
Ref angle: $\scriptsize \theta ={{30}^\circ}$
Sine is positive in the first and second quadrants.
$\scriptsize \theta ={{180}^\circ}-{{30}^\circ}={{150}^\circ}$
General solution: $\scriptsize \theta ={{30}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{150}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
$\scriptsize \sin \theta =1$:
Ref angle: $\scriptsize \theta ={{90}^\circ}$
General solution: $\scriptsize \theta ={{90}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$
3. .
\scriptsize \displaystyle \begin{align*}4\sin x+3\tan x & =\displaystyle \frac{3}{{\cos x}}+4\quad \quad \cos x\ne 0\therefore x\ne {{90}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\\\therefore 4\sin x+3\displaystyle \frac{{\sin x}}{{\cos x}}-\displaystyle \frac{3}{{\cos x}}-4 & =0\\\therefore 4\sin x\cos x+3\sin x-3-4\cos x & =0\\\therefore 4\cos x(\sin x-1)+3(\sin x-1) & =0\\\therefore (\sin x-1)(4\cos x+3) & =0\\\therefore \sin x=1\text{ } & \text{or }\cos x=-\displaystyle \frac{3}{4}\end{align*}
$\scriptsize \sin x=1$:
Ref angle: $\scriptsize x={{90}^\circ}$
General solution: $\scriptsize x={{90}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
$\scriptsize \cos x=-\displaystyle \frac{3}{4}$:
Ref angle: $\scriptsize x={{138.59}^\circ}$
Cosine is negative in the second and third quadrants.
$\scriptsize x={{360}^\circ}-{{138.59}^\circ}={{221.41}^\circ}$
General solution: $\scriptsize x={{138.59}^\circ}+k{{.360}^\circ}\text{ or }x={{221.49}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
For the interval $\scriptsize \left[ {{{0}^\circ},{{{180}}^\circ}} \right]$: $\scriptsize x={{90}^\circ}\text{ or }x={{138.59}^\circ}$
4. .
\scriptsize \begin{align*}6\cos \theta -5 & =\displaystyle \frac{4}{{\cos \theta }}\quad \quad \cos \theta \ne 0\therefore \theta \ne {{90}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\\\therefore 6{{\cos }^{2}}\theta -5\cos \theta -4 & =0\\\therefore (3\cos \theta -4)(2\cos \theta +1) & =0\\\therefore \cos \theta =\displaystyle \frac{4}{3}\text{ } & \text{or }\cos \theta =-\displaystyle \frac{1}{2}\end{align*}
$\scriptsize \cos \theta =\displaystyle \frac{4}{3}$ – No solution
$\scriptsize \cos \theta =-\displaystyle \frac{1}{2}$:
Ref angle: $\scriptsize \theta ={{120}^\circ}$
Cosine is negative in the second and third quadrants.
$\scriptsize \theta ={{360}^\circ}-({{120}^\circ})={{240}^\circ}$
General solution: $\scriptsize \theta ={{120}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{240}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
5. .
\scriptsize \begin{align*}\cos 2x-\cos x+1 & =0\\\therefore 2{{\cos }^{2}}x-1-\cos x+1 & =0\\\therefore 2{{\cos }^{2}}x-\cos x & =0\\\therefore \cos x(2\cos x-1) & =0\\\therefore \cos x=0\text{ } & \text{or }\cos x=\displaystyle \frac{1}{2}\end{align*}
$\scriptsize \cos x=0$:
Ref angle: $\scriptsize x={{90}^\circ}$
General solution: $\scriptsize x={{90}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$
$\scriptsize \cos x=\displaystyle \frac{1}{2}$
Ref angle: $\scriptsize \theta ={{60}^\circ}$
Cosine is positive in the first and fourth quadrants.
$\scriptsize \theta ={{360}^\circ}-{{60}^\circ}={{300}^\circ}$
General solution: $\scriptsize \theta ={{60}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{300}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
For the interval $\scriptsize \left[ {{{0}^\circ},{{{360}}^\circ}} \right]$: $\scriptsize x=\text{6}{{\text{0}}^\circ}\text{ or }x={{90}^\circ}\text{ or }x={{180}^\circ}\text{ or }x=\text{30}{{\text{0}}^\circ}\text{ or }x=\text{36}{{\text{0}}^\circ}$
2. $\scriptsize \cos 2\theta =2{{\cos }^{2}}\theta -1$
1. .
\scriptsize \begin{align*}\text{LHS}=\cos 2\theta +3\cos \theta -1\\=2{{\cos }^{2}}\theta -1+3\cos \theta -1\\=2{{\cos }^{2}}\theta +3\cos \theta -2=\text{RHS}\end{align*}.
2. .
\scriptsize \begin{align*}\cos 2x-3\cos x-1 & =0\\\therefore 2{{\cos }^{2}}x-3\cos x-2 & =0\\\therefore (2\cos x+1)(\cos x-2) & =0\\\therefore \cos x=-\displaystyle \frac{1}{2}\text{ } & \text{or }\cos x=2\end{align*}
$\scriptsize \cos \theta =-\displaystyle \frac{1}{2}$:
Ref angle: $\scriptsize \theta ={{120}^\circ}$
Cosine is negative in the second and third quadrants.
$\scriptsize \theta ={{360}^\circ}-({{120}^\circ})={{240}^\circ}$
General solution: $\scriptsize \theta ={{120}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{240}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}$
$\scriptsize \cos x=2$ – No solution
For the interval $\scriptsize \left[ {{{0}^\circ},{{{360}}^\circ}} \right]$: $\scriptsize \theta ={{120}^\circ}\text{ or }\theta ={{240}^\circ}$
3. .
\scriptsize \begin{align*}3\cos 2x-\sin 2x-1 & =0\\\therefore 3({{\cos }^{2}}x-{{\sin }^{2}}x)-2\sin x\cos x-({{\sin }^{2}}x+{{\cos }^{2}}) & =0\\\therefore 3{{\cos }^{2}}-3{{\sin }^{2}}x-2\sin x\cos x-{{\sin }^{2}}x-{{\cos }^{2}}x & =0\\\therefore 2{{\cos }^{2}}x-2\cos x\sin x-4{{\sin }^{2}}x= & 0\\\therefore {{\cos }^{2}}x-\cos x\sin x-2{{\sin }^{2}}x & =0\\\therefore (\cos x+\sin x)(\cos x-2\sin x) & =0\\\therefore \cos x=-\sin x\text{ } & \text{or }\cos x=2\sin x\end{align*}
\scriptsize \begin{align*}\cos x & =-\sin x\\\therefore 1 & =-\displaystyle \frac{{\sin x}}{{\cos x}}\\\therefore \tan x & =-1\end{align*}
Ref angle: $\scriptsize x=-{{45}^\circ}$
General solution: $\scriptsize x={{135}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$
\scriptsize \begin{align*}\cos x & =2\sin x\\\therefore 1 & =2\cdot \displaystyle \frac{{\sin x}}{{\cos x}}\\\therefore 2\tan x & =1\\\therefore \tan x & =\displaystyle \frac{1}{2}\end{align*}
Ref angle: $\scriptsize x={{26.6}^\circ}$
General solution: $\scriptsize x={{26.6}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}$
For the interval $\scriptsize \left[ {{{0}^\circ},{{{360}}^\circ}} \right]$: $\scriptsize x={{26.6}^\circ}\text{ or }x={{135}^\circ}\text{ or }x={{206.6}^\circ}\text{ or }x={{315}^\circ}$

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