Unit outcomes

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

• Solve trigonometric equations using a general solution.

What you should know

Before you start this unit, make sure you can:

• Define the three basic trigonometric ratios of sine, cosine and tangent.
• Solve basic linear equations.
• Calculate with the special angles. Refer to unit 1 of this subject outcome if you need help with this.
• Use the trig reduction formulae. Refer to unit 2 of this subject outcome if you need help with this.
• Draw and work with the CAST diagram to determine in which quadrants each of the trig ratios are positive or negative. Refer to level 2 subject outcome 3.6 unit 1 if you need help with this.

Introduction

If [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex], what is [latex]\scriptsize \theta[/latex]? If you said that [latex]\scriptsize \theta ={{30}^\circ}[/latex] because of your knowledge of special angles from unit 1, you would be absolutely correct and yet still not completely correct! How can this be?

Remember that the trig functions are periodic – they repeat themselves over and over again. We know already, for example, that [latex]\scriptsize \sin {{150}^\circ}=\sin ({{180}^\circ}-{{30}^\circ})=\sin {{30}^\circ}=\displaystyle \frac{1}{2}[/latex] as well. But the function repeats itself again every [latex]\scriptsize {{360}^\circ}[/latex]. Therefore, there are infinitely many solutions to [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex].

If you have an internet connection, visit this simulation to see all the solutions to [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex] for the interval [latex]\scriptsize -{{720}^\circ}\le \theta \le {{720}^\circ}[/latex].

You will find that there are eight in total.

All of these solutions are shown in Figure 1.

Therefore, it is not good enough to say that the solution to [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex] is [latex]\scriptsize \theta ={{30}^\circ}[/latex]. Nor can we list all the possible solutions. We need a way to give a general solution that can generate all the possible solutions for [latex]\scriptsize \theta \in \mathbb{R}\text{ }[/latex].

The general solution for [latex]\scriptsize \sin \theta[/latex]

Let’s look at the solutions for [latex]\scriptsize \theta[/latex] for [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex] for the interval [latex]\scriptsize -{{720}^\circ}\le \theta \le {{720}^\circ}[/latex] again.

[latex]\scriptsize -{{690}^\circ},-{{570}^\circ},-{{330}^\circ},-{{210}^\circ},{{30}^\circ},{{150}^\circ},{{390}^\circ},{{510}^\circ}[/latex]. Can you see any pattern? Look back at Figure 1 to help you.

What if you arrange the solutions in two columns like this:
[latex]\scriptsize \begin{align*} -{{690}^\circ}&&-{{570}^\circ}\\ -{{330}^\circ}&&-{{210}^\circ}\\ {{30}^\circ}&&{{150}^\circ}\\ {{390}^\circ}&&{{510}^\circ} \end{align*}[/latex]

Can you see the pattern now? Hopefully you can see that each solution in each column is separated from the next by [latex]\scriptsize {{360}^\circ}[/latex], the period of the sine function. Look at the first column again. If we use [latex]\scriptsize {{30}^\circ}[/latex] (the answer that a calculator would give us) as the starting angle or the reference angle, look at how we can generate all the other solutions.
[latex]\scriptsize \begin{align*}&{{30}^\circ}-2\times {{360}^\circ}-{{690}^\circ}\\&{{30}^\circ}-1\times {{360}^\circ}=-{{330}^\circ}\\&{{30}^\circ}\\&{{30}^\circ}+1\times {{360}^\circ}={{390}^\circ}\end{align*}[/latex]

We can do the same for the second column of solutions using [latex]\scriptsize {{180}^\circ}-{{30}^\circ}={{150}^\circ}[/latex] as the starting angle.
[latex]\scriptsize \begin{align*}&{{150}^\circ}-2\times {{360}^\circ}=-{{570}^\circ}\\&{{150}^\circ}-1\times {{360}^\circ}=-{{210}^\circ}\\&{{150}^\circ}\\&{{150}^\circ}+1\times {{360}^\circ}={{510}^\circ}\end{align*}[/latex]

So, with just the two starting angles [latex]\scriptsize {{30}^\circ}[/latex] and [latex]\scriptsize {{150}^\circ}[/latex], we can generate every other solution to [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex] by adding integer ([latex]\scriptsize \mathbb{Z}[/latex]) multiples of the period of the sine function i.e. [latex]\scriptsize {{360}^\circ}[/latex].

We write the full or general solution to [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex] as follows:
[latex]\scriptsize \theta ={{30}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{150}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]

Take note!

We call the answer we get for [latex]\scriptsize \theta[/latex], (usually on a calculator) and from which we generate all the other possible solutions, the reference angle.

Example 4.1

Determine the general solution for [latex]\scriptsize \sin \theta =0.35[/latex].

Solution

Step 1: Use a calculator to determine the reference angle
[latex]\scriptsize \begin{align*}\sin \theta &=0.35\\\therefore \theta &={{20.5}^\circ}\end{align*}[/latex]
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
[latex]\scriptsize \sin \theta =0.35[/latex]. 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 [latex]\scriptsize \theta ={{20.5}^\circ}[/latex]). We need to find the second quadrant solution. We know that [latex]\scriptsize \sin ({{180}^\circ}-\theta )=\sin \theta[/latex]. Therefore, the second quadrant solution is [latex]\scriptsize {{180}^\circ}-{{20.5}^\circ}={{159.5}^\circ}[/latex].

Step 3: Generate the general solution
[latex]\scriptsize \theta ={{20.5}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{159.5}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]

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 [latex]\scriptsize k[/latex].

[latex]\scriptsize k=-2[/latex]:
[latex]\scriptsize \begin{align*}&\theta ={{20.5}^\circ}-2\times {{360}^\circ}\text{ or }\theta ={{159.5}^\circ}-2\times {{360}^\circ}\\&\therefore \theta =-{{699.5}^\circ}\text{ or }\theta =-{{560.5}^\circ}\end{align*}[/latex]
[latex]\scriptsize \sin (-{{699.5}^\circ})=0.35[/latex]
[latex]\scriptsize \sin (-{{560.5}^\circ})=0.35[/latex]

Our general solution is correct.

Example 4.2

Solve for [latex]\scriptsize \theta[/latex] if [latex]\scriptsize 4\sin \theta =-3[/latex] for the interval [latex]\scriptsize [-{{180}^\circ},{{180}^\circ}][/latex].

Solution

[latex]\scriptsize \begin{align*}4\sin \theta =-3&\\\therefore \sin \theta =-\displaystyle \frac{3}{4}&=-0.75\end{align*}[/latex]

Step 1: Use a calculator to determine the reference angle
When we have a negative ratio, we ignore the sign when finding the reference angle.
[latex]\scriptsize \begin{align*}&\sin \theta =0.75\\&\therefore \theta ={{48.6}^\circ}\end{align*}[/latex]

Step 2: Use the CAST diagram to determine any other possible solutions
Our equation is [latex]\scriptsize \sin \theta =-0.75[/latex]. In other words, sine is negative. Sine is negative in the third and fourth quadrants. Our reference angle is [latex]\scriptsize \theta ={{48.6}^\circ}[/latex].

Third quadrant: [latex]\scriptsize \sin ({{180}^\circ}+\theta )=-\sin \theta[/latex]. Therefore, the third quadrant solution is [latex]\scriptsize {{180}^\circ}+{{48.6}^\circ}={{228.6}^\circ}[/latex].

Fourth quadrant: [latex]\scriptsize \sin ({{360}^\circ}-\theta )=-\sin \theta[/latex]. Therefore, the fourth quadrant solution is [latex]\scriptsize {{360}^\circ}-{{48.6}^\circ}={{311.4}^\circ}[/latex].

Step 3: Generate the general solution
[latex]\scriptsize \theta ={{228.6}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{311.4}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]

Step 4: Generate the solution for the specific range
In this instance, we were not asked for the general solution but only for the solutions in [latex]\scriptsize [-{{180}^\circ},{{180}^\circ}][/latex]. We need to use our general solution to generate solutions that fall within this range by adding or subtracting multiples of [latex]\scriptsize {{360}^\circ}[/latex].

[latex]\scriptsize \theta ={{228.6}^\circ}-1\times {{360}^\circ}=-{{131.4}^\circ}\text{or }\theta ={{311.4}^\circ}-1\times {{360}^\circ}=-{{48.6}^\circ}[/latex]

Step 5: Check your general solution
It is always a good idea to check that your final solutions satisfy the original equation.
[latex]\scriptsize \sin (-{{131.4}^\circ})=-0.75[/latex]
[latex]\scriptsize \sin (-{{48.6}^\circ})=-0.75[/latex]

Exercise 4.1

1. Determine the general solution for [latex]\scriptsize 2\sin \theta =\sqrt{3}[/latex].
2. Solve for [latex]\scriptsize x[/latex] if [latex]\scriptsize 5\sin x=-2[/latex] and [latex]\scriptsize {{0}^\circ}\le x\le {{360}^\circ}[/latex].

The full solutions are at the end of the unit.

The general solution for [latex]\scriptsize \cos \theta[/latex] and [latex]\scriptsize \tan \theta[/latex]

The general solution for [latex]\scriptsize \cos \theta =y[/latex] is basically the same as that for sine. Remember that cosine also has a period of [latex]\scriptsize {{360}^\circ}[/latex]. The only difference occurs when you find the quadrants in which cosine is either positive or negative.

The general solution for [latex]\scriptsize \tan \theta =y[/latex] is basically the same as that for sine except that the period of tangent is [latex]\scriptsize {{180}^\circ}[/latex]. The other difference occurs when you find the quadrants in which tangent is either positive or negative.

Example 4.3

Determine the general solution for [latex]\scriptsize 3\cos x=\sin {{14}^\circ}[/latex].

Solution

[latex]\scriptsize \begin{align*}3\cos x&=\sin {{14}^\circ}\\\therefore \cos x&=\displaystyle \frac{{\sin {{{14}}^\circ}}}{3}\end{align*}[/latex]

Step 1: Use a calculator to determine the reference angle
[latex]\scriptsize \begin{align*}\cos x & =\displaystyle \frac{{\sin {{{14}}^\circ}}}{3}\\\therefore x & ={{85.4}^\circ}\end{align*}[/latex]

Step 2: Use the CAST diagram to determine any other possible solutions
Our equation is [latex]\scriptsize \cos x=\displaystyle \frac{{\sin {{{14}}^\circ}}}{3}[/latex]. Because [latex]\scriptsize \sin {{14}^\circ} \gt 0[/latex] we know that [latex]\scriptsize \cos x \gt 0[/latex]. Cosine is positive in the first and fourth quadrants. Our reference angle is [latex]\scriptsize \theta ={{85.4}^\circ}[/latex].

First quadrant: [latex]\scriptsize \theta ={{85.4}^\circ}[/latex]

Fourth quadrant: [latex]\scriptsize \cos ({{360}^\circ}-\theta )=\cos \theta[/latex]
[latex]\scriptsize {{360}^\circ}-{{85.4}^\circ}={{274.6}^\circ}[/latex].

Step 3: Generate the general solution
[latex]\scriptsize \theta ={{85.4}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{274.6}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]

Step 4: Check your general solution
[latex]\scriptsize k=2[/latex]:
[latex]\scriptsize \theta ={{85.4}^\circ}+{{2.360}^\circ}={{805.4}^\circ}\text{ or }\theta ={{274.6}^\circ}+{{2.360}^\circ}={{994.6}^\circ}[/latex]
[latex]\scriptsize \cos {{805.4}^\circ}=\displaystyle \frac{{\sin {{{14}}^\circ}}}{3}[/latex]
[latex]\scriptsize \sin {{994.6}^\circ}=\displaystyle \frac{{\sin {{{14}}^\circ}}}{3}[/latex]

Example 4.4

Solve for [latex]\scriptsize \alpha[/latex] if [latex]\scriptsize \tan \alpha =-5.2[/latex] and [latex]\scriptsize {{0}^\circ}\le \alpha \le {{360}^\circ}[/latex].

Solution

Step 1: Use a calculator to determine the reference angle
[latex]\scriptsize \begin{align*}\tan \alpha&=5.2&&\text{Remember, when finding the reference angle we always use the positive ratio}\\\therefore \alpha&={{79.1}^\circ}\end{align*}[/latex]

Step 2: Use the CAST diagram to determine any other possible solutions
Our equation is [latex]\scriptsize \tan \alpha =-5.2[/latex]. [latex]\scriptsize \tan \alpha \lt 0[/latex].

Second quadrant: [latex]\scriptsize \tan ({{180}^\circ}-\theta )=-\tan \theta[/latex]. Therefore, the second quadrant solution is [latex]\scriptsize {{180}^\circ}-{{79.1}^\circ}={{100.9}^\circ}[/latex].

Fourth quadrant: [latex]\scriptsize \tan ({{360}^\circ}-\theta )=-\tan \theta[/latex]. Therefore, the fourth quadrant solution is [latex]\scriptsize {{360}^\circ}-{{79.1}^\circ}={{280.9}^\circ}[/latex].

Step 3: Generate the general solution
[latex]\scriptsize \alpha ={{100.9}^\circ}+k{{.180}^\circ}\text{ or }\alpha ={{280.9}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}[/latex]

Note: Because the period of tangent is [latex]\scriptsize {{180}^\circ}[/latex], the general solution includes integer multiples of [latex]\scriptsize {{180}^\circ}[/latex]. Therefore, it is also possible to generate the general solution from only one angle. [latex]\scriptsize {{280.9}^\circ}={{100.9}^\circ}+1\times {{180}^\circ}[/latex].

The simplest general solution is thus [latex]\scriptsize \alpha ={{100.9}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}[/latex]

Step 4: Generate the solution for the specific range
In this instance, we were not asked for the general solution but only for the solutions in[latex]\scriptsize {{0}^\circ}\le \alpha \le {{360}^\circ}[/latex]. We need to use our general solution to generate solutions that fall within this range by adding or subtracting multiples of [latex]\scriptsize {{180}^\circ}[/latex].

[latex]\scriptsize \alpha ={{100.9}^\circ}+0\times {{180}^\circ}={{100.9}^\circ}\text{or }\alpha ={{100.9}^\circ}+1\times {{180}^\circ}={{280.9}^\circ}[/latex]

Step 5: Check your general solutions
[latex]\scriptsize \tan {{100.9}^\circ}=-5.2[/latex]
[latex]\scriptsize \tan {{280.9}^\circ}=-5.2[/latex]

Exercise 4.2

1. Determine the general solution for [latex]\scriptsize 5-3\tan \theta =0[/latex].
2. Solve for [latex]\scriptsize x[/latex] if [latex]\scriptsize 3\cos x-1=-2[/latex] for the interval [latex]\scriptsize [{{0}^\circ},{{360}^\circ}][/latex].

The full solutions are at the end of the unit.

Solve more complicated trig equations

In level 3 subject outcome 2.1 units 7, 9 and 11, we learnt about the functions [latex]\scriptsize y=a\sin kx[/latex], [latex]\scriptsize y=a\cos kx[/latex] and [latex]\scriptsize y=a\tan kx[/latex]. We saw, for example, that the period of the function [latex]\scriptsize y=\sin 2x[/latex] was no longer [latex]\scriptsize {{360}^\circ}[/latex]. The period was actually [latex]\scriptsize \displaystyle \frac{{{{{360}}^\circ}}}{2}={{180}^\circ}[/latex].

How many solutions are there to [latex]\scriptsize \sin x=0.5[/latex]? How many solutions are there to [latex]\scriptsize \sin 2x=0.5[/latex]?
Figure 2 shows the solutions to [latex]\scriptsize \sin x=0.5[/latex]. As you would expect, these are [latex]\scriptsize x={{30}^\circ}[/latex] and [latex]\scriptsize x={{150}^\circ}[/latex] for the interval [latex]\scriptsize [{{0}^\circ},{{360}^\circ}][/latex].

Now look at the solutions to [latex]\scriptsize \sin 2x=0.5[/latex] in Figure 3. There are three important things to notice.

1. The period of the graph has halved and the graph repeats twice in the interval [latex]\scriptsize [{{0}^\circ},{{360}^\circ}][/latex]. Therefore, there are double the number of solutions in the same interval.
2. The solution [latex]\scriptsize x={{30}^\circ}[/latex] from before is now [latex]\scriptsize x=\displaystyle \frac{{{{{30}}^\circ}}}{2}={{15}^\circ}[/latex]. The solution [latex]\scriptsize x={{150}^\circ}[/latex] from before is now [latex]\scriptsize x=\displaystyle \frac{{{{{150}}^\circ}}}{2}={{75}^\circ}[/latex].
3. Solutions to [latex]\scriptsize \sin x=0.5[/latex] are separated by integer multiples of [latex]\scriptsize {{360}^\circ}[/latex]. Solutions to [latex]\scriptsize \sin 2x=0.5[/latex] are seperated by interger multiples of [latex]\scriptsize \displaystyle \frac{{{{{360}}^\circ}}}{2}={{180}^\circ}[/latex].

Therefore, when we solve equations such as [latex]\scriptsize \sin 2\theta =\displaystyle \frac{1}{2}[/latex], we need to modify the general solution to take into account the period of the function is now [latex]\scriptsize {{180}^\circ}[/latex].

So, if the general solution to [latex]\scriptsize \sin \theta =\displaystyle \frac{1}{2}[/latex] is [latex]\scriptsize \theta ={{30}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{150}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex], then the general solution to [latex]\scriptsize \sin 2\theta =\displaystyle \frac{1}{2}[/latex] is [latex]\scriptsize \theta ={{15}^\circ}+k{{.180}^\circ}\text{ or }\theta ={{75}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }[/latex].

The same is true for cosine and tangent equations.

Example 4.5

Determine the general solution for [latex]\scriptsize \theta[/latex] if [latex]\scriptsize 5\sin 2\theta =3[/latex].

Solution

[latex]\scriptsize \begin{align*}5\sin 2\theta & =3\\\therefore \sin 2\theta & =\displaystyle \frac{3}{5}\end{align*}[/latex]

Step 1: Use a calculator to determine the reference angle
[latex]\scriptsize \begin{align*}\sin 2\theta & =\displaystyle \frac{3}{5}\\\therefore 2\theta & ={{36.9}^\circ}\end{align*}[/latex]

Note: We keep working with the reference angle as [latex]\scriptsize 2\theta[/latex] until we generate the general solution.

Step 2: Use the CAST diagram to determine any other possible solutions
Our equation is [latex]\scriptsize \sin 2\theta =\displaystyle \frac{3}{5}[/latex]. [latex]\scriptsize \sin 2\theta \gt 0[/latex]. Sine is positive in the first and second quadrants.

First quadrant: [latex]\scriptsize 2\theta ={{36.9}^\circ}[/latex]

Second quadrant: [latex]\scriptsize \sin ({{180}^\circ}-\theta )=\sin \theta[/latex]
[latex]\scriptsize 2\theta ={{180}^\circ}-{{36.9}^\circ}={{143.2}^\circ}[/latex].

Step 3: Generate the general solution
[latex]\scriptsize \begin{align*}2\theta & ={{36.9}^\circ}+k{{.360}^\circ}\text{ or 2}\theta ={{143.2}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }\\\therefore \theta & ={{18.45}^\circ}+k{{.180}^\circ}\text{ or }\theta ={{71.6}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }\end{align*}[/latex]

Step 4: Check your general solution
[latex]\scriptsize k=2[/latex]:
[latex]\scriptsize \therefore \theta ={{18.45}^\circ}+2\times {{180}^\circ}={{378.45}^\circ}\text{ or }\theta ={{71.6}^\circ}+2\times {{180}^\circ}={{431.6}^\circ}[/latex]
[latex]\scriptsize \sin (2\times {{378.45}^\circ})=0.6[/latex]
[latex]\scriptsize \sin (2\times {{431.6}^\circ})=0.6[/latex]

Example 4.6

Solve for [latex]\scriptsize \theta[/latex] if [latex]\scriptsize \tan (3\theta -{{38}^\circ})=-5[/latex] for the interval [latex]\scriptsize [-{{180}^\circ},{{180}^\circ}][/latex].

Solution

Step 1: Use a calculator to determine the reference angle
[latex]\scriptsize \begin{align*}\tan (3\theta -{{38}^\circ}) & =5\quad \text{Remember, when finding the reference angle, use the positive ratio}\\\therefore 3\theta -{{38}^\circ} & ={{78.7}^\circ}\quad \text{Keep your reference angle in terms of }3\theta -{{38}^\circ}\end{align*}[/latex]

Note: We keep working with the reference angle as [latex]\scriptsize 3\theta -{{38}^\circ}[/latex] until we generate the general solution.

Step 2: Use the CAST diagram to determine any other possible solutions
Our equation is [latex]\scriptsize \tan (3\theta -{{38}^\circ})=-5[/latex]. [latex]\scriptsize \tan (3\theta -{{38}^\circ}) \lt 0[/latex]. Tangent is negative in the second and fourth quadrants.

Second quadrant: [latex]\scriptsize \tan ({{180}^\circ}-\theta )=-\tan \theta[/latex]
[latex]\scriptsize {{180}^\circ}-{{78.7}^\circ}={{101.3}^\circ}[/latex]

Step 3: Generate the general solution
[latex]\scriptsize \begin{align*}3\theta -{{38}^\circ} & ={{101.3}^\circ}+k{{.180}^\circ},\text{ }k\in \mathbb{Z}\quad \text{Add }{{38}^\circ}\text{to the angle}\\\therefore 3\theta & ={{139.3}^\circ}+k{{.180}^\circ},\text{ }k\in \mathbb{Z}\quad \text{Divide both the angle and the period by }3\\\therefore \theta &={{46.37}^\circ}+k{{.60}^\circ},\text{ }k\in \mathbb{Z}\end{align*}[/latex]

Step 4: Generate the solution for the specific range
The interval is [latex]\scriptsize [-{{180}^\circ},{{180}^\circ}][/latex].
[latex]\scriptsize \theta ={{46.37}^\circ}-3\times {{60}^\circ}=-{{133.63}^\circ}[/latex] or
[latex]\scriptsize \theta ={{46.37}^\circ}-2\times {{60}^\circ}=-{{73.63}^\circ}[/latex] or
[latex]\scriptsize \theta ={{46.37}^\circ}-1\times {{60}^\circ}=-{{13.63}^\circ}[/latex] or
[latex]\scriptsize \theta ={{46.37}^\circ}+0\times {{60}^\circ}={{46.37}^\circ}[/latex] or
[latex]\scriptsize \theta ={{46.37}^\circ}+1\times {{60}^\circ}={{106.37}^\circ}[/latex] or
[latex]\scriptsize \theta ={{46.37}^\circ}+2\times {{60}^\circ}={{166.37}^\circ}[/latex]

Exercise 4.3

1. Determine the general solution for [latex]\scriptsize \cos \left( {\displaystyle \frac{1}{2}x} \right)=0.54[/latex].
2. Solve for [latex]\scriptsize x[/latex] if [latex]\scriptsize 2\sin \left( {\displaystyle \frac{{3x}}{2}+{{{10}}^\circ}} \right) =-1[/latex] for the interval [latex]\scriptsize [-{{180}^\circ},{{180}^\circ}][/latex].

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

• How to find the general solution for trig equations involving sine, cosine and tangent and simple angles.
• How to find the general solution for trig equations involving sine, cosine and tangent and more complex angles.
• How to find the solutions to trig equations involving sine, cosine and tangent for specific intervals.

Unit 4: Assessment

Suggested time to complete: 45 minutes

All the questions below have been adapted from NC(V) examination questions.

1. Determine the general solution for [latex]\scriptsize \sin \theta +1=0.3[/latex].
2. Calculate the value/s of [latex]\scriptsize \theta[/latex] if [latex]\scriptsize 2\tan \theta =0.279[/latex] where [latex]\scriptsize \theta \in [{{0}^\circ},{{360}^\circ}][/latex].
3. Calculate the value/s of [latex]\scriptsize \theta[/latex] if [latex]\scriptsize 5\cos 2\theta =-2.7[/latex] where [latex]\scriptsize \theta \in [{{0}^\circ},{{360}^\circ}][/latex].
4. Calculate the value/s of [latex]\scriptsize \theta[/latex] if [latex]\scriptsize \tan (3\theta -{{48}^\circ})=3.2[/latex] where [latex]\scriptsize \theta \in [{{0}^\circ},{{180}^\circ}][/latex].
5. Determine the value/s of [latex]\scriptsize \theta[/latex] in the following equation without a calculator if [latex]\scriptsize {{0}^\circ}\le \theta \le {{360}^\circ}[/latex].
   [latex]\scriptsize \sin \theta +\sqrt{2}=-\sin \theta[/latex]

The full solutions are at the end of the unit.

Unit 4: Solutions

Exercise 4.1

1. .
   [latex]\scriptsize \begin{align*}2\sin \theta & =\sqrt{3}\\\therefore \sin \theta & =\displaystyle \frac{{\sqrt{3}}}{2}\end{align*}[/latex]
   Ref angle: [latex]\scriptsize \theta ={{60}^\circ}[/latex] (you should have recognised this as a special angle ratio)
   [latex]\scriptsize \sin \theta \gt 0[/latex] in first and second quadrant. Therefore, [latex]\scriptsize \theta ={{60}^\circ}\text{ or }\theta ={{180}^\circ}-{{60}^\circ}={{120}^\circ}[/latex].
   General solution: [latex]\scriptsize \theta ={{60}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{120}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]
2. .
   [latex]\scriptsize \begin{align*}5\sin x & =-2\\\therefore \sin x & =-\displaystyle \frac{2}{5}\end{align*}[/latex]
   Ref angle: [latex]\scriptsize x={{23.6}^\circ}[/latex]
   [latex]\scriptsize \sin x \lt 0[/latex] in third and fourth quadrants.
   [latex]\scriptsize \theta ={{180}^\circ}+23.6={{203.6}^\circ}\text{ or }\theta ={{360}^\circ}-{{23.6}^\circ}={{336.4}^\circ}[/latex].
   General solution: [latex]\scriptsize \theta ={{203.6}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{336.4}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]
   Specific solution for [latex]\scriptsize {{0}^\circ}\le x\le {{360}^\circ}[/latex]: [latex]\scriptsize \theta ={{203.6}^\circ}\text{ or }\theta ={{336.4}^\circ}[/latex]

Back to Exercise 4.1

Exercise 4.2

1. .
   [latex]\scriptsize \begin{align*}5-3\tan \theta & =0\\\therefore \tan \theta & =\displaystyle \frac{5}{3}\end{align*}[/latex]
   Ref angle: [latex]\scriptsize \theta ={{59}^\circ}[/latex]
   [latex]\scriptsize \tan \theta \gt 0[/latex] in the first and third quadrants.
   [latex]\scriptsize \theta ={{59}^\circ}\text{ or }\theta ={{180}^\circ}+{{59}^\circ}={{239}^\circ}[/latex]
   General solution: [latex]\scriptsize \theta ={{59}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }[/latex]
2. .
   [latex]\scriptsize \begin{align*}3\cos x-1 & =-2\\\therefore 3\cos x & =-1\\\therefore \cos x & =-\displaystyle \frac{1}{3}\end{align*}[/latex]
   Ref angle: [latex]\scriptsize x={{70.5}^\circ}[/latex]
   [latex]\scriptsize \cos x \lt 0[/latex] in the second and third quadrants.
   [latex]\scriptsize x={{180}^\circ}-{{70.5}^\circ}={{109.5}^\circ}\text{ or }x={{180}^\circ}+{{70.5}^\circ}={{250.5}^\circ}[/latex]
   General solution: [latex]\scriptsize x={{109.5}^\circ}+k{{.360}^\circ}\text{ or }x={{250.5}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]
   Specific solution for [latex]\scriptsize [{{0}^\circ},{{360}^\circ}][/latex]: [latex]\scriptsize x={{109.5}^\circ}\text{ or }x={{250.5}^\circ}[/latex]

Back to Exercise 4.2

Exercise 4.3

1. .
   [latex]\scriptsize \begin{align*}\cos \left( {\displaystyle \frac{1}{2}x} \right) & =0.54\\\therefore \displaystyle \frac{1}{2}x & ={{57.3}^\circ}\end{align*}[/latex]
   Ref angle: [latex]\scriptsize \displaystyle \frac{1}{2}x={{57.3}^\circ}[/latex]
   [latex]\scriptsize \cos \left( {\displaystyle \frac{1}{2}x} \right) \gt 0[/latex] in the first and fourth quadrants.
   [latex]\scriptsize \displaystyle \frac{1}{2}x={{57.3}^\circ}\text{ or }\displaystyle \frac{1}{2}x={{360}^\circ}-{{57.3}^\circ}={{302.7}^\circ}[/latex]
   General solution:
   [latex]\scriptsize \begin{align*}\displaystyle \frac{1}{2}x & ={{57.3}^\circ}+k{{.360}^\circ}\text{ or }\displaystyle \frac{1}{2}x={{302.7}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }\\\therefore x & ={{114.6}^\circ}+k{{.720}^\circ}\text{ or }x={{605.4}^\circ}+k{{.720}^\circ},k\in \mathbb{Z}\text{ }\end{align*}[/latex]
2. .
   [latex]\scriptsize \begin{align*}2\sin \left( {\displaystyle \frac{{3x}}{2}+{{{10}}^\circ}} \right) & =-1\\\therefore \sin \left( {\displaystyle \frac{{3x}}{2}+{{{10}}^\circ}} \right) &=-\displaystyle \frac{1}{2}\end{align*}[/latex]
   Ref angle: [latex]\scriptsize \left( {\displaystyle \frac{{3x}}{2}+{{{10}}^\circ}} \right) ={{30}^\circ}[/latex]
   [latex]\scriptsize \sin \left( {\displaystyle \frac{{3x}}{2}+{{{10}}^\circ}} \right) \lt 0[/latex] in the third and fourth quadrants.
   [latex]\scriptsize \displaystyle \frac{{3x}}{2}+{{10}^\circ}={{180}^\circ}+{{30}^\circ}={{210}^\circ}\text{ or }\displaystyle \frac{{3x}}{2}+{{10}^\circ}={{360}^\circ}-{{30}^\circ}={{330}^\circ}[/latex]
   General solution:
   [latex]\scriptsize \begin{align*}\displaystyle \frac{{3x}}{2}+{{10}^\circ} & ={{210}^\circ}+k{{.360}^\circ}\text{ or }\displaystyle \frac{{3x}}{2}+{{10}^\circ}={{330}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }\\\therefore \displaystyle \frac{{3x}}{2} & ={{200}^\circ}+k{{.360}^\circ}\text{ or }\displaystyle \frac{{3x}}{2}={{320}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }\\\therefore x & ={{133.33}^\circ}+k{{.240}^\circ}\text{ or }\displaystyle \frac{{3x}}{2}={{213.33}^\circ}+k{{.240}^\circ},k\in \mathbb{Z}\text{ }\end{align*}[/latex]
   Specific solution for [latex]\scriptsize [-{{180}^\circ},{{180}^\circ}][/latex]:
   [latex]\scriptsize x ={{133.33}^\circ}-1\times {{240}^\circ}=-{{106.67}^\circ}[/latex] or
   [latex]\scriptsize x ={{133.33}^\circ}+0\times {{240}^\circ}={{133.33}^\circ}[/latex] or
   [latex]\scriptsize x ={{213.33}^\circ}-1\times {{240}^\circ}=-{{26.67}^\circ}[/latex]

Back to Exercise 4.3

Unit 4: Assessment

1. .
   [latex]\scriptsize \begin{align*}\sin \theta +1 & =0.3\\\therefore \sin \theta & =-0.7\end{align*}[/latex]
   Ref angle: [latex]\scriptsize \theta ={{44.4}^\circ}[/latex]
   [latex]\scriptsize \sin \theta \lt 0[/latex]: [latex]\scriptsize \theta ={{180}^\circ}+{{44.4}^\circ}={{224.4}^\circ}\text{ or }\theta ={{360}^\circ}-{{44.4}^\circ}={{315.6}^\circ}[/latex]
   General solution: [latex]\scriptsize \theta ={{224.4}^\circ}+k\text{.36}{{\text{0}}^\circ}\text{ or }\theta ={{315.6}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }[/latex]
2. .
   [latex]\scriptsize \begin{align*}2\tan\theta&=0.279\\\therefore \tan\theta&=0.1395\end{align*}[/latex]
   Ref angle: [latex]\scriptsize \theta ={{8.0}^\circ}[/latex]
   [latex]\scriptsize \tan \theta \gt 0:[/latex] [latex]\scriptsize \theta ={{8.0}^\circ}\text{ or }\theta ={{180}^\circ}+{{8.0}^\circ}={{188}^\circ}[/latex]
   General solution: [latex]\scriptsize \theta ={{8.0}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }[/latex]
   Specific solution for [latex]\scriptsize \theta \in [{{0}^\circ},{{360}^\circ}][/latex]:
   [latex]\scriptsize \theta ={{8}^\circ}+0\times {{180}^\circ}={{8}^\circ}[/latex] or
   [latex]\scriptsize \theta ={{8.0}^\circ}+1\times {{180}^\circ}={{188.0}^\circ}[/latex]
3. .
   [latex]\scriptsize \begin{align*}5\cos2\theta&=-2.7\\\therefore \cos2\theta&=-0.54\end{align*}[/latex]
   Ref angle: [latex]\scriptsize 2\theta ={{57.3}^\circ}[/latex]
   [latex]\scriptsize \cos 2\theta \lt 0[/latex]: [latex]\scriptsize 2\theta ={{180}^\circ}-{{57.3}^\circ}={{122.7}^\circ}\text{ or }2\theta ={{180}^\circ}+{{57.3}^\circ}={{237.3}^\circ}[/latex]
   General solution:
   [latex]\scriptsize \begin{align*}2\theta ={{122.7}^\circ}+k{{.360}^\circ}\text{ or }2\theta ={{237.3}^\circ}+k{{.360}^\circ},k\in \mathbb{Z}\text{ }\\\therefore \theta ={{61.35}^\circ}+k{{.180}^\circ}\text{ or }\theta ={{118.65}^\circ}+{{180}^\circ},k\in \mathbb{Z}\text{ }\end{align*}[/latex]
   Specific solution for [latex]\scriptsize \theta \in [{{0}^\circ},{{360}^\circ}][/latex]:
   [latex]\scriptsize \theta ={{61.35}^\circ}+0\times {{180}^\circ}={{61.35}^\circ}[/latex] or
   [latex]\scriptsize \theta ={{61.35}^\circ}+1\times {{180}^\circ}={{241.35}^\circ}[/latex] or
   [latex]\scriptsize \theta ={{118.65}^\circ}+0\times {{180}^\circ}={{118.65}^\circ}[/latex] or
   [latex]\scriptsize \theta ={{118.65}^\circ}+1\times {{180}^\circ}={{298.65}^\circ}[/latex]
4. [latex]\scriptsize \tan (3\theta -{{48}^\circ})=3.2[/latex]
   Ref angle: [latex]\scriptsize 3\theta -{{48}^\circ}={{72.6}^\circ}[/latex]
   [latex]\scriptsize 3\theta -{{48}^\circ}={{72.6}^\circ}\text{ or }3\theta -{{48}^\circ}={{180}^\circ}+{{72.6}^\circ}={{252.6}^\circ}[/latex]:
   General solution:
   [latex]\scriptsize \begin{align*}3\theta -{{48}^\circ} & ={{72.6}^\circ}+k{{.180}^\circ}\text{ or }3\theta -{{48}^\circ}={{252.6}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }\\\therefore 3\theta & ={{120.6}^\circ}+k{{.180}^\circ}\text{ or }3\theta ={{300.6}^\circ}+k{{.180}^\circ},k\in \mathbb{Z}\text{ }\\\therefore \theta & ={{40.2}^\circ}+k{{.60}^\circ}\text{ or }\theta ={{100.2}^\circ}+k{{.60}^\circ},k\in \mathbb{Z}\text{ }\end{align*}[/latex]
   But [latex]\scriptsize {{40.2}^\circ}+1\times {{60}^\circ}={{100.2}^\circ}[/latex]
   Therefore, simplest general solution is [latex]\scriptsize \theta ={{40.2}^\circ}+k{{.60}^\circ},k\in \mathbb{Z}\text{ }[/latex]
   Specific solution for [latex]\scriptsize \theta \in [{{0}^\circ},{{180}^\circ}][/latex]:
   [latex]\scriptsize \theta ={{40.2}^\circ}+0\times {{60}^\circ}={{40.2}^\circ}[/latex] or
   [latex]\scriptsize \theta ={{40.2}^\circ}+1\times {{60}^\circ}={{100.2}^\circ}[/latex] or
   [latex]\scriptsize \theta ={{40.2}^\circ}+2\times {{60}^\circ}={{160.2}^\circ}[/latex]
5. .
   [latex]\scriptsize \begin{align*}\sin \theta +\sqrt{2} & =-\sin \theta \\\therefore 2\sin \theta & =-\sqrt{2}\\\therefore \sin \theta & =-\displaystyle \frac{{\sqrt{2}}}{2}=-\displaystyle \frac{{\sqrt{2}}}{{\sqrt{2}\times \sqrt{2}}}=-\displaystyle \frac{1}{{\sqrt{2}}}\end{align*}[/latex]
   Ref angle: [latex]\scriptsize \theta ={{45}^\circ}[/latex]
   [latex]\scriptsize \sin \theta \lt 0[/latex]: [latex]\scriptsize \theta ={{180}^\circ}+{{45}^\circ}={{225}^\circ}\text{ or }\theta ={{360}^\circ}-{{45}^\circ}={{315}^\circ}[/latex]
   General solution: [latex]\scriptsize \theta ={{225}^\circ}+k{{.360}^\circ}\text{ or }\theta ={{315}^\circ}+k{{.360}^\circ}[/latex]
   Specific solution for [latex]\scriptsize {{0}^\circ}\le \theta \le {{360}^\circ}[/latex]:
   [latex]\scriptsize \theta ={{225}^\circ}+{{0.360}^\circ}={{225}^\circ}[/latex] or
   [latex]\scriptsize \theta ={{315}^\circ}+{{0.360}^\circ}={{315}^\circ}[/latex]

Back to Unit 4: Assessment