The idea, step by step
The calculator gives a starting angle
Inverse sine, cosine or tangent returns one principal value. The equation may have other solutions in the given interval. Use a unit-circle sketch or graph to find the matching quadrants, and list only angles inside the stated bounds.
For sin x = 1/2 from 0° to 360°, the reference angle is 30°. Sine is positive in quadrants I and II, giving 30° and 150°.
Solve the inside angle first
For cos(2x) = 1/2 with 0° ≤ x ≤ 360°, the inside angle 2x ranges from 0° to 720°. Find every solution in that expanded interval, then divide by 2.
Use exact values when possible and substitute each candidate into the original equation. Include endpoints when the interval uses ≤; do not include an undefined tangent value.
A common trap
Finding solutions for 2x only between 0° and 360° when x extends to 360°.
Instead: Transform the interval along with the angle expression before listing solutions.
See the thinking
Find both quadrants
Solve sin x = 1/2 for 0° ≤ x ≤ 360°.
- The reference angle is .
- Sine is positive in quadrants I and II.
- or .
Two complete cycles
Solve cos(2x) = 1/2 for 0° ≤ x ≤ 360°.
- The inside angle ranges from 0° to 720°.
- Cosine equals 1/2 at 60°, 300°, 420° and 660° in that interval.
- Divide every inside-angle solution by 2.
Check your understanding
Solve cos x = −1 for 0° ≤ x ≤ 360°.
Show answer
180°
The leftmost unit-circle point has cosine −1.
Your turn
6 questions · 15 marks · Work at your own pace.
Read the questions online
- Solve sin x = √3/2 for 0° ≤ x ≤ 360°.2 marks
- Solve cos x = −1/2 for 0° ≤ x ≤ 360°.2 marks
- Solve tan x = 1 for 0° ≤ x ≤ 360°.2 marks
- Solve sin(2x) = 0 for 0° ≤ x ≤ 360°.3 marks
- Solve 2 cos x + 1 = 0 for 0° ≤ x ≤ 360°.3 marks
- A student solves sin x = −1/2 on 0° ≤ x ≤ 360° and gives only 330°. 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 = √3/2 for 0° ≤ x ≤ 360°.2 marks
Reference angle is 60°; sine is positive in I and II.
60°, 120°. - Solve cos x = −1/2 for 0° ≤ x ≤ 360°.2 marks
Reference angle is 60°; cosine is negative in II and III.
120°, 240°. - Solve tan x = 1 for 0° ≤ x ≤ 360°.2 marks
Tangent has period 180° and is positive in I and III.
45°, 225°. - Solve sin(2x) = 0 for 0° ≤ x ≤ 360°.3 marks
For 0° ≤ 2x ≤ 720°, zeros are 0°, 180°, 360°, 540° and 720°.
Divide by 2.
0°, 90°, 180°, 270°, 360°. - Solve 2 cos x + 1 = 0 for 0° ≤ x ≤ 360°.3 marks
.
Cosine is −1/2 at 120° and 240°.
120°, 240°. - A student solves sin x = −1/2 on 0° ≤ x ≤ 360° and gives only 330°. Find the missing solution and explain.3 marks
Negative sine occurs in quadrants III and IV.
The reference angle is 30°, so 180° + 30° = 210°.
210° is also a solution.
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 ↗