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1.5 Trigonometry

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Question 17
i.

Solve, for 0<x<π0 < x < \pi0<x<π, the equation

(4x−5)(4sin⁡2x−3)=0 (4x - 5)(4\sin^2 x - 3) = 0 (4x−5)(4sin2x−3)=0

giving your answers in terms of π\piπ where appropriate.

[3]
ii.

Solve, for 0<θ<360∘0 < \theta < 360^\circ0<θ<360∘, the equation

11cos⁡θ=3cos⁡2θ+8 11\cos \theta = 3\cos 2\theta + 8 11cosθ=3cos2θ+8

giving your answers to one decimal place.

[5]
Markscheme

1.5 Trigonometry Questions

  1. A Level
  2. /Maths
  3. /1.5 Trigonometry

97 exam-style questions on WJEC A Level Maths 1.5 Trigonometry, covering 1.5.1 Trigonometry, 1.5.2 Trigonometry, 1.5.3 Trigonometry, 1.5.4 Trigonometry, and 1.5.5 Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.

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