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0606/23 May/Jun 2025, Question 12: Solving Trig Equations

Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 10 marks

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The question, in short

For f(x) = 15cos²(3x + 1.5) + 7sin(3x + 1.5) − 13 on −0.3 ≤ x ≤ 0.5 radians, find where f(x) = 0 and locate the two stationary points of y = f(x).

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Part (a) · 5 marks

Determine the values of x in the given interval that satisfy f(x) = 0.

  1. 115(1 − sin²(3x + 1.5)) + 7sin(3x + 1.5) − 13 = 0
  2. 215sin²(3x + 1.5) − 7sin(3x + 1.5) − 2 = 0
  3. 3(5sin(3x + 1.5) + 1)(3sin(3x + 1.5) − 2) = 0
  4. 40.6 ≤ 3x + 1.5 ≤ 3: sin = −1/5 gives no value; sin = 2/3 gives 3x + 1.5 = 0.7297 or 2.4119
  5. 5x = −0.257 or x = 0.304

Answerx = −0.257, x = 0.304

Part (b) · 5 marks

Determine the x-values where the gradient of y = f(x) is zero, for the two stationary points.

  1. 1f′(x) = 30cos(3x + 1.5)(−3sin(3x + 1.5)) + 21cos(3x + 1.5)
  2. 2f′(x) = 3cos(3x + 1.5)(7 − 30sin(3x + 1.5)) = 0
  3. 3cos(3x + 1.5) = 0: 3x + 1.5 = π/2, x = 0.0236
  4. 4sin(3x + 1.5) = 7/30: in range 3x + 1.5 = π − 0.2355 = 2.906, x = 0.469

Answerx = 0.0236, x = 0.469

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Also searched as: 0606/23/M/J/25 · 0606_s25_qp_23 · 0606 s25 23 · question 12.
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