Solve trigonometric equations between 0° and 360°
12th grade (Year 13 in England) · 16 questions · Common Core: HSF.TF.B.7 · England: A-level Pure Mathematics
Name
Date
Score/ 16
Question 1
Solve sin x = 0.1, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 2
Solve cos x = √3/2 for 0° ≤ x < 360°.
- A. 210°, 330°
- B. 30°, 210°
- C. 30°, 330°
- D. 30°, 150°
Question 3
Solve sin x = 0.5, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 4
Solve cos x = −√3/2 for 0° ≤ x < 360°.
- A. 30°, 150°
- B. 150°, 210°
- C. 30°, 330°
- D. 30°, 210°
Question 5
Solve sin x = 0.6, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 6
Solve tan x = −√3 for 0° ≤ x < 360°.
- A. 60°, 120°
- B. 240°, 300°
- C. 120°, 300°
- D. 120°, 240°
Question 7
Solve sin x = 0.1, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 8
Solve cos x = −√3/2 for 0° ≤ x < 360°.
- A. 150°, 210°
- B. 210°, 330°
- C. 150°, 330°
- D. 30°, 330°
Question 9
Solve sin x = 0.1, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 10
Solve cos x = √2/2 for 0° ≤ x < 360°.
- A. 45°, 135°
- B. 45°, 225°
- C. 135°, 225°
- D. 45°, 315°
Question 11
Solve sin x = 0.1, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 12
Solve sin x = −1/2 for 0° ≤ x < 360°.
- A. 150°, 330°
- B. 30°, 150°
- C. 30°, 210°
- D. 210°, 330°
Question 13
Solve sin x = 0.4, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 14
Solve cos x = √3/2 for 0° ≤ x < 360°.
- A. 30°, 210°
- B. 30°, 330°
- C. 150°, 210°
- D. 30°, 150°
Question 15
Solve sin x = 0.1, giving the solution between 90° and 180° to 1 decimal place.
Answer
Question 16
Solve cos x = −1/2 for 0° ≤ x < 360°.
- A. 120°, 240°
- B. 60°, 120°
- C. 60°, 240°
- D. 60°, 300°
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