Suitable trigonometric equations on a stated interval | Year 10

01 / LEARN

Understand the idea.

The calculator gives a starting angle

An inverse trigonometric function returns one principal angle, not every solution. Use the unit circle or a graph to identify all matching quadrants, then keep only angles inside the question's stated interval.

For example, positive sine corresponds to points above the horizontal axis. A horizontal line through the unit circle usually meets it twice, so the same positive sine can belong to an acute and an obtuse angle.

Solve the inside angle first

When the angle is an expression such as $2 x$ , transform the interval as well as the equation. If $0^{\circ} \leq x \leq 360^{\circ}$ , then $0^{\circ} \leq 2 x \leq 720^{\circ}$ , covering two complete cycles.

List all inside-angle solutions first, then undo the expression to recover $x$ . Check endpoints when they are included and substitute candidates into the original equation; tangent is undefined wherever cosine is zero.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find both quadrants

Solve sin x = 1/2 for 0° ≤ x ≤ 360°.

Worked solution

1
Find the reference angle.
$sin 30^{\circ} = \frac{1}{2}$
The special triangle gives an exact acute angle.
2
Use the positive-sine quadrants.
$x = 30^{\circ} or x = 180^{\circ} - 30^{\circ} = 150^{\circ}$
Sine is positive in quadrants I and II.
3
Check the interval and endpoints.
$sin 0^{\circ} = sin 360^{\circ} = 0$
Both candidates lie in the interval; neither included endpoint supplies an additional solution.

Answer

30°, 150°

EXAMPLE 02

Two complete cycles

Solve cos(2x) = 1/2 for 0° ≤ x ≤ 360°.

Worked solution

1
Transform the interval.
$u = 2 x, 0^{\circ} \leq u \leq 720^{\circ}$
The inside angle runs twice as far as $x$ .
2
Find the first-cycle angles.
$cos u = \frac{1}{2} \Rightarrow u = 60^{\circ}, 300^{\circ}$
Cosine is positive in quadrants I and IV.
3
Include the second cycle.
$u = 60^{\circ}, 300^{\circ}, 420^{\circ}, 660^{\circ}$
Adding $360^{\circ}$ gives all further matches within the expanded interval; its endpoints have cosine $1$ .
4
Divide every inside angle by two.
$x = 30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}$
Each solution of $u = 2 x$ must be converted back to the requested variable.

Answer

30°, 150°, 210°, 330°

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve cos x = −1 for 0° ≤ x ≤ 360°.

Show answer & reasoning

180°

Cosine is the horizontal coordinate. It equals $- 1$ only at the leftmost unit-circle point, so $x = 180^{\circ}$ is the only solution in the stated interval.

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Solve $sin x = \frac{\sqrt{3}}{2}$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
2 marks
Solve $cos x = - \frac{1}{2}$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
2 marks
Solve $tan x = 1$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
2 marks
Solve $sin (2 x) = 0$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
3 marks
Solve $2 cos x + 1 = 0$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
3 marks
A student solves $sin x = - \frac{1}{2}$ on $0^{\circ} \leq x \leq 360^{\circ}$ and gives only $330^{\circ}$ . Find the missing solution and explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $sin x = \frac{\sqrt{3}}{2}$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
2 marks
Reference angle is $60^{\circ}$ ; sine is positive in I and II.
$60^{\circ}$ , $120^{\circ}$ .
Solve $cos x = - \frac{1}{2}$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
2 marks
Reference angle is $60^{\circ}$ ; cosine is negative in II and III.
$120^{\circ}$ , $240^{\circ}$ .
Solve $tan x = 1$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
2 marks
Tangent has period $180^{\circ}$ and is positive in I and III.
$45^{\circ}$ , $225^{\circ}$ .
Solve $sin (2 x) = 0$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
3 marks
For $0^{\circ} \leq 2 x \leq 720^{\circ}$ , zeros are $0^{\circ}$ , $180^{\circ}$ , $360^{\circ}$ , $540^{\circ}$ and $720^{\circ}$ .
Divide by 2.
$0^{\circ}$ , $90^{\circ}$ , $180^{\circ}$ , $270^{\circ}$ , $360^{\circ}$ .
Solve $2 cos x + 1 = 0$ for $0^{\circ} \leq x \leq 360^{\circ}$ .
3 marks
$cos x = - \frac{1}{2}$ .
Cosine is $- \frac{1}{2}$ at $120^{\circ}$ and $240^{\circ}$ .
$120^{\circ}$ , $240^{\circ}$ .
A student solves $sin x = - \frac{1}{2}$ on $0^{\circ} \leq x \leq 360^{\circ}$ and gives only $330^{\circ}$ . Find the missing solution and explain.
3 marks
Negative sine occurs in quadrants III and IV.
The reference angle is $30^{\circ}$ , so $180^{\circ} + 30^{\circ} = 210^{\circ}$ .
$210^{\circ}$ is also a solution.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian school course scope

Right-triangle applications include AC9M10M03. The approved chapter also develops non-right-triangle methods, exact values and trigonometric functions within one normal Year 10 course.

Official curriculum source ↗

In this lesson

Understand the idea.

The calculator gives a starting angle

Solve the inside angle first

See the method in action.

Find both quadrants

Find the reference angle.

Use the positive-sine quadrants.

Check the interval and endpoints.

Two complete cycles

Transform the interval.

Find the first-cycle angles.

Include the second cycle.

Divide every inside angle by two.

Try it yourself.

Check your understanding

Keep practising