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WMA12/01 Oct 2023, Question 3: Harder Trig Equations

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 7 marks

The question, in short

Solve a trigonometric equation involving tan and sin for angles between 0 and 360 degrees, then use that result to find the smallest positive angle satisfying a related equation with 2x + 40°.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 5 marks

Find all angles θ in 0 < θ ≤ 360° satisfying 2 tan θ + 3 sin θ = 0, giving answers to one decimal place where needed.

  1. 1Use tan θ = sin θ/cos θ: 2 sin θ/cos θ + 3 sin θ = 0
  2. 2Multiply by cos θ: 2 sin θ + 3 sin θ cos θ = 0, so sin θ(2 + 3 cos θ) = 0
  3. 3sin θ = 0 ⇒ θ = 180°, 360°
  4. 42 + 3 cos θ = 0 ⇒ cos θ = −2/3
  5. 5θ = 131.8° or 360° − 131.8° = 228.2°

Answerθ = 131.8°, 180°, 228.2°, 360°

Part (b) · 2 marks

Using part (a) or another method, determine the smallest positive x for which 2 tan(2x + 40°) + 3 sin(2x + 40°) = 0, to one decimal place.

  1. 1Since 2x + 40° > 40°, the smallest value of 2x + 40° from part (a) is 131.8°
  2. 22x + 40° = 131.8°
  3. 32x = 91.8°, so x = 45.9°

Answerx = 45.9°

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Also searched as: WMA12 October 2023 · WMA12/01 · wma12 oct23 · IAS Maths Pure Mathematics 2 October 2023 · question 3.
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