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Grade 11 · Trigonometry
Trig Equations & General Solutions (Grade 11)
MEMORANDUM
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Class:
Date:
Mark
/ 22
Answer all questions. Show all your working — marks are awarded for method as well as the final answer. Teacher copy: accept any correct equivalent method.
- 1Determine the general solution of sinθ = 0,5(3)θ = 30° + k·360° or θ = 150° + k·360°, k ∈ ℤ✓✓✓ (3)
- 2Determine the general solution of cosθ = −0,5(3)ref = 60°, cos negative in QII and QIII →θ = ±120° + k·360°, k ∈ ℤ✓✓✓ (3)
- 3Determine the general solution of tanθ = √3(2)θ = 60° + k·180°, k ∈ ℤ✓✓ (2)
- 4Solve sinθ = 0,5 for θ ∈ [0° ; 360°](2)θ = 30° or θ = 150°✓✓ (2)
- 5Solve cosθ = 0,7660 for θ ∈ [0° ; 360°] (round to one decimal)(3)ref = 40,0° → θ = 40,0° or θ = 320,0°✓✓✓ (3)
- 6Determine the general solution of tan(2x − 30°) = 1(3)2x − 30° = 45° + k·180° → x = 37,5° + k·90°, k ∈ ℤ✓✓✓ (3)
- 7State the reference angle for sinθ = 0,8660.(1)ref = sin−1(0,8660) = 60° (to the nearest degree)✓ (1)
- 8Determine the general solution of sin 2θ = cosθ.(4)cosθ = sin(90° − θ), so sin 2θ = sin(90° − θ) (1)
2θ = 90° − θ + k·360° ⇒ 3θ = 90° + k·360° ⇒ θ = 30° + k·120° (1)
or 2θ = 180° − (90° − θ) + k·360° ⇒ θ = 90° + k·360° (1)
θ = 30° + k·120° or θ = 90° + k·360°, k ∈ ℤ (1)✓✓✓✓ (4) - 9Why must k ∈ ℤ be written down?(1)It states that the pattern repeats for every integer k — without it the solution is incomplete and loses a mark.✓ (1)
TOTAL: 22 marks