Year 10A - Solving trigonometric equations - Free practice

Year 10A Solving Trigonometric Equations Worksheets & Practice

Find all solutions over a given interval, with full working shown every time, free.

Quick answer

How do you solve trigonometric equations?

To solve a trigonometric equation such as sin(x) = k, find the reference angle using the inverse trig function, then use the unit circle's symmetry and periodicity to find all solutions in the given interval. For example, sin(x) = 0.5 gives x = 30° and x = 150° in [0°, 360°].

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Worked examples

Year 10A solving trigonometric equations: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Solving a basic sine equation
Solve $\sin(x) = \dfrac{\sqrt{3}}{2}$ for $0° \leq x \leq 360°$ .
Show worked solution
Key idea: Find the reference angle from exact values, then use the unit circle to identify all angles in the given interval where sine has that value.
1. Recall the exact value: $\sin(60°) = \dfrac{\sqrt{3}}{2}$ , so the reference angle is $60°$ .
2. Sine is positive in the first and second quadrants.
3. First quadrant solution: $x = 60°$ .
4. Second quadrant solution: $x = 180° - 60° = 120°$ .
5. Both values lie in $[0°, 360°]$ , so both are valid.
Answer: $x = 60°$ or $x = 120°$
Example 2 · Solving a cosine equation requiring rearrangement
Solve $2\cos(x) + \sqrt{3} = 0$ for $0° \leq x \leq 360°$ .
Show worked solution
Key idea: Rearrange the equation to isolate $\cos(x)$ first, then use the unit circle and symmetry to find all solutions in the interval.
1. Rearrange: $2\cos(x) = -\sqrt{3}$ , so $\cos(x) = -\dfrac{\sqrt{3}}{2}$ .
2. Recall the exact value: $\cos(30°) = \dfrac{\sqrt{3}}{2}$ , so the reference angle is $30°$ .
3. Cosine is negative in the second and third quadrants.
4. Second quadrant solution: $x = 180° - 30° = 150°$ .
5. Third quadrant solution: $x = 180° + 30° = 210°$ .
Answer: $x = 150°$ or $x = 210°$
Example 3 · Solving a trigonometric equation in a real-life context
The depth of water (in metres) at a pier is modelled by $d = 5\sin(\theta) + 7$ , where $\theta$ is an angle parameter that increases from $0°$ to $360°$ over one tidal cycle. A boat needs a minimum depth of $9.5$ metres to dock safely. Find all values of $\theta$ in $[0°, 360°]$ for which the depth is exactly $9.5$ metres.
Show worked solution
Key idea: Substitute the required depth into the model and solve the resulting trigonometric equation over the given interval.
1. Set $d = 9.5$ : $5\sin(\theta) + 7 = 9.5$ .
2. Rearrange: $5\sin(\theta) = 2.5$ , so $\sin(\theta) = 0.5$ .
3. Recall the exact value: $\sin(30°) = 0.5$ , so the reference angle is $30°$ .
4. Sine is positive in the first and second quadrants.
5. First quadrant: $\theta = 30°$ . Second quadrant: $\theta = 180° - 30° = 150°$ .
Answer: $\theta = 30°$ or $\theta = 150°$

Common mistakes

Where students usually lose marks

Forgetting the second solution

$\sin(x) = \frac{1}{2}$ , so $x = 30°$ only.

Every basic sine or cosine equation has two solutions in $[0°, 360°]$ (unless the value is $\pm 1$ or $0$ ). Always check both quadrants where the ratio has the correct sign using the unit circle.

Mixing up the quadrant rules for sine, cosine and tangent

Cosine is negative, so the solutions are in the first and second quadrants.

Use the CAST rule: All ratios are positive in Q1, Sine only in Q2, Tangent only in Q3, Cosine only in Q4. Cosine is negative in Q2 and Q3, not Q1 and Q2.

Using the wrong formula for the third- or fourth-quadrant angle

Reference angle is $45°$ , third quadrant solution: $x = 360° - 45° = 315°$ .

The third quadrant solution is $180°$ plus the reference angle ( $180° + 45° = 225°$ ), and the fourth quadrant solution is $360°$ minus the reference angle ( $360° - 45° = 315°$ ). Make sure you use the right formula for each quadrant.

Forgetting to rearrange before finding the reference angle

$2\sin(x) = 1$ , reference angle $= \sin^{-1}(2) =$ error.

Always isolate the trigonometric ratio completely before finding the reference angle. Divide both sides by the coefficient first to get $\sin(x) = \frac{1}{2}$ , then find the reference angle.

Try one yourself

Your turn

Find all solutions to $\cos(x) = -\dfrac{\sqrt{3}}{2}$ in the interval $0^\circ \leq x \leq 360^\circ$ . Enter both solutions separated by a comma (smaller value first). Give each answer in degrees as an exact whole number.

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About this worksheet

Why we built it

Solving trigonometric equations is a key skill in the Year 10A extension strand and a direct stepping stone to senior Methods maths. These worksheets guide students from reading exact values off the unit circle through to finding all solutions of equations like sin(x) = 0.5 or 2cos(x) + 1 = 0 over a specified interval, with fully worked solutions so students can check every line of their reasoning.

What's covered

Sub-skills your student will practise

✓Reading exact trigonometric values from the unit circle
✓Solving equations of the form sin(x) = k and cos(x) = k
✓Solving equations of the form tan(x) = k
✓Finding all solutions over a given interval in degrees or radians
✓Using symmetry and periodicity to locate multiple solutions
✓Solving simple equations that require rearrangement first

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Frequently asked

Questions parents ask about solving trigonometric equations

What does 'solve a trigonometric equation' actually mean?

It means finding all angle values that make the equation true over a given interval. For example, solving sin(x) = 0.5 for 0 to 360 degrees gives x = 30° and x = 150°, because sine equals 0.5 at both those angles.

Do students need to know radians for Year 10A trig equations?

Yes. The 10A extension content expects students to work in both degrees and radians. Radians are introduced here and are essential for senior Mathematical Methods (Mathematics Advanced in NSW), so it is worth building comfort with both units early.

Why do trigonometric equations have more than one solution?

Because sine, cosine and tangent are periodic, meaning their values repeat at regular intervals. The unit circle makes this visual: two different angles can share the same sine or cosine value within a single 360° cycle, and more solutions appear in wider intervals.

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