A Level Maths, A Level Trigonometry, KS5

Solving Trigonometric Equations – A Level Maths Revision

Solving trigonometric equations

There aren’t any complicated equations when it comes to sourcing your A Level Maths revision – simply sticking with Beyond will equal a great learning experience as proved with this post on solving trigonometric equations.

What are Trigonometric Equations?

A trigonometric equation is an equation where a trigonometric function has been applied to the variable.

For the most part, they are solved like a normal equation – do the inverse of each operation to both sides of the equation, in the correct order, until you have the variable on it’s on just one side of the equals sign.

The difference is, unlike operations such as addition or multiplication, trigonometric functions are not one-to-one – they are periodic. This is something you must take into account when solving trigonometric equations.

This revision blog will run through the process of solving trigonometric equations, with a few practice questions and answers at the end.

If you want to see the information contained, with more questions, in PDF or PowerPoint form, click here. If you want to practise some of the prior-knowledge from GCSE, try this multiple-choice quiz. Once you’re ready, try these exam-style questions on all AS-level trigonometry.

You should already know how to use the three trigonometric ratios – sine, cosine and tangent – to find missing angles or sides in a triangle using the graphs for each of the ratios.

The graphs are periodic, which means that they repeat after a certain interval. The sin and cos functions have a period of 360° and the tan function has a period of 180°.

Sine Graph

Sine graph

Cosine Graph

Cosine graph

Tangent Graph

Tangent graph

This means that there is often more than one solution to a trigonometric equation.

For example, if cosx = 1, x could be 0° or 360° or -360°. There’s an infinite amount of solutions to the equation. In this case, you can keep adding or subtracting 360° to get a new solution. To limit this, there is usually an interval in which you need to find all the solutions, for example 0° ≤ x ≤ 360°.

This doesn’t always necessarily mean that there is a unique solution in that interval. Usually, you will still need to find equivalent solutions.

In some cases, there are no solutions to trigonometric equations, for example sinx = 2 has no solutions since sinx only takes values between -1 and 1.

If you consider sinx = 0.5, you can use the graphs of y = sinx and y = 0.5 to look for solutions in the range -360° ≤ x ≤ 360°.

Solving trigonometric equations - sinx = 2

You can see that there are 4 solutions in this interval. A calculator will give the first one, x = 30° and you can use the symmetry of the graph to find the others.

Between x = 0° and x = 180°, the graph of y = sinx is symmetrical about x = 90°. The first result, x = 30°, is 60° less than 90° therefore the next result with be 60° greater than 90°, x = 150°.

To find the others, use the fact that y = sinx is periodic, with a period of 360°: you can simply take 360° from each of the solutions.

This means the four solutions are x = -330°, -210°, 30° and 150°, for -360° ≤ x ≤ 360°.

Solving trigonometric equations is just one topic with which we lend our helping hand. You can find a greater range of supportive A Level Maths resources right here:

A Level Maths resources

Solving Trigonometric Equations – Example Question 1

Solve sinθ = 0.668 for 0° ≤ θ ≤ 720°, giving your answer in degrees correct to 1 decimal place.

The symbol θ is a Greek letter and is pronounced theta. It is commonly used to represent an angle.

Start by finding the base solution – the one your calculator gives you:

\theta = sin^{-1} (0.668) = 41.9^{\circ} \text{ (1d.p.)}

Then, consider the sine graph to help you find the other solutions. You can see from the graph that there are 4 solutions in the given interval.

Solving trigonometric equations - example question 1

Use the symmetry of the graph to find the second positive solution. You can see that solution 2 is the same distance from 180° as solution 1 is from 0°, so solution 2 is:

180 - 41.9 = 138.1^{\circ} \text{ (1d.p.)}

You can then use the periodic nature of the graph to find solutions 3 and 4. Since the graph repeats every 360°, you can add 360° to the existing solutions to find others in the given interval. So, solutions 3 and 4 are:

41.9 + 360 = 401.9^{\circ} \text{ (1d.p.)}\\138.1 + 360 = 498.1^{\circ} \text{ (1d.p.)}

Therefore, the solutions in the interval 0° ≤ θ ≤ 720° are:

\theta = 41.9^{\circ}, 138.1^{\circ}, 401.9^{\circ}, 498.1^{\circ} \text{ (1d.p.)}

Example Question 2

Solve \boldsymbol{2tanx=3} for 0° ≤ x ≤ 360°, giving your answer in degrees, correct to 3 significant figures.

First, we need to rearrange this equation to make tanx the subject:

2\tan{x}=3 \\ \tan{x}=1.5

As in example 1, start by finding the base solution:

x = \tan{^{-1}}1.5 = 56.3^{\circ} \text{ (3s.f.)}

Then, consider the graph of y = \tan{x} to find the other solutions. There are 2 solutions in the given interval:

Example question 2

Unlike the sine and cosine graph, the tangent graph has a period of 180°, so you can simply add or subtract 180° to the base solution. Solution 2 is:

56.3 + 180 = 236^{\circ} \text{ (3s.f.)}

The two solutions for 2tanx = 3 , where 0° ≤ x ≤ 360° are:

x = 56.3^{\circ} \text{ and } 236^{\circ} \text{ (3s.f.)}

Example Question 3

Solve \boldsymbol{5\cos{y} + 7 = 10} for -360° ≤ y ≤ 360°, giving your answer correct to 1 decimal place.

Start by rearranging the equation so it is in a similar for to the previous questions:

5\cos{y} + 7 = 10 \\ 5\cos{y} = 3 \\ \cos{y} = 0.6

Next, find the base solution:

y = \cos{^{-1}}(0.6) = 53.1^{\circ}

Then, consider the cosine graph for the given interval to look for other solutions:

Example question 3

Use the symmetry of the graph to find the second positive solution. You can see that solution 2 is the same distance from 360° as solution 1 is from 0°, so solution 2 is:

360-53.1=306.9^{\circ} \text{ (1d.p.)}

You can then use the periodic nature of the graph to find solutions 3 and 4. Since the graph repeats every 360°, you can add or subtract 360° from your existing solutions to find others in the given interval. Therefore, solutions 3 and 4 are:

53.1 -360 =-306.9^{\circ} \text{ (1d.p.) and } \\306.9-360=-53.1^{\circ} \text{ (1d.p.)}

And the solutions, in the interval -360° ≤ y ≤ 360°, are:

y = -306.9^{\circ}, -53.1^{\circ}, 53.1^{\circ} \text{ and } 306.9^{\circ} \text{ (1d.p.)}

Summary:
You can use the symmetry of the trigonometry graphs to find additional solutions for trigonometric equations. If the base solution is x:

● for sine, the second solution will be 180 – x
● for cosine, the second solution will be 360 – x
● for tangent, the second solution will be 180 + x

In addition to adding or subtracting 360° for sine or cosine, or 180° for tan.

Example Question 4

Solve \boldsymbol{\sin{(2x)} = -0.7} for 0° ≤ x ≤ 360°, giving your answer to a suitable degree of accuracy.

This is slightly different to the previous questions – you will need to use the graph of y = \sin{(2x)} rather than simply y = \sin{x} . This halves the period of the graph, which means there will be 4 solutions in the given interval.

Example question 4

You can simplify this by substituting 2x for another variable, such as \theta . If you define \theta = 2x you then need to solve the equation \sin{\theta} = -0.7 .

You will also need to adjust your interval. If \theta = 2x , then x = \frac{1}{2}\theta . This means the interval becomes:

0^{\circ} \leq \frac{1}{2} \theta \leq 360^{\circ}

or:

0^{\circ} \leq \theta \leq 720^{\circ}

Therefore, you need to solve sinθ = -0.7 for 0° ≤ θ ≤ 720°.

Start by finding the base solution:

\sin{}^{-1}(-0.7) = -44.4

Note that, in this case, the base solution is outside of the interval, but you can continue
in the same method to find 4 solutions within the interval.

Subtract the base solution from 180° to find the second solution:

180 - - 44.4 = 224.4^{\circ}

Then add 360° to each of the 2 solutions to find the others:

360 + -44.4 = 315.6^{\circ} \\ 360 + 224.4 = 584.4^{\circ}

You now have four solutions, but only three are within the given interval. To get the fourth solution, add add 360° to 315.6°. Don’t add it to 584.4° as this will take you above 720°.

360 + 315.6 = 675.6^{\circ}

The solutions for \theta are 224.4°, 315.6°, 584.4° and 675.6°. Since x = \frac{1}{2}\theta , we need to halve each of these to find the four values of x in the range 0° ≤ x ≤ 360°:

x = 112^{\circ}, 158^{\circ}, 292^{\circ} \text{ and } 338^{\circ} \text{ (3s.f.)}

Example Question 5

Solve the equation \boldsymbol{4\cos{(3x-15)}-1=1} for -180° ≤ x ≤ 180°, giving your answer to a suitable degree of accuracy.

First, make \cos{(3x-15)} the subject:

4\cos{(3x-15)}-1 = 1 \\ 4\cos{(3x-15)}=2 \\ \cos{(3x-15)}=0.5

Next, substitute 3x - 15 for θ in both the equation and interval. If \theta = 3x - 15 then rearranging to make x the subject gives x = \frac{\theta + 15}{3} . The interval becomes:

-180^{\circ} \leq \frac{\theta + 15}{3} \leq 180^{\circ} \\ -540^{\circ} \leq \theta + 15 \leq 540^{\circ} \\ -555^{\circ} \leq \theta \leq 525^{\circ}

Therefore, you need to solve \cos{\theta} = 0.5 for -555° ≤ θ ≤ 525°:

\theta = \cos{}^{-1}(0.5) = 60^{\circ} \\ 360 - 60 = 300^{\circ}

Therefore, 60° and 300° are the first two solutions.

Now, simply add 360° to each:

60 + 360 = 420^{\circ} \\ 300 + 360 = 660^{\circ}

660° is outside the interval so you can ignore it. You also need to subtract 360° from the first two solutions:

60 - 360 = -300^{\circ} \\ 300 - 360 = -60^{\circ}

There is one more solution to be found. Do this by subtracting 360° from -60°.

-60 - 360 = -420^{\circ}

The means the solutions for θ are: -420°, -300°, -60°, 60°, 300° and 420°.

Using x = \frac{\theta + 15}{3} you can find the 6 solutions for x :

x = -135^{\circ}, -95^{\circ}, -15^{\circ}, 25^{\circ}, 105^{\circ} \text{ and }145^{\circ}

Solving Trigonometric Equations – Practice Questions

Here’s some to try yourself. Solve each of the trigonometric equations for 0° ≤ y ≤ 180°.

a. \sin{y} = 0.3

b. \tan{y} = 2

c. \sqrt{2}\cos{y} = 3

d. 5\sin{y} + 1 = 3

e. \tan{(3y+20)}=1.5

Answers

a. y = 17.5° and 162° (3s.f.)

b. y = 63.4° (3s.f.)

c. y = 30°

d. y = 23.6° and 156.4°

e. y = 12.1°, 72.1° and 132.1°

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