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Trigonometric Identities and Equations

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Question 19
a.

Show that the equation

2sin⁡2x=4cos⁡2x−cos⁡x 2\sin^2 x = 4\cos^2 x - \cos x 2sin2x=4cos2x−cosx

can be expressed in the form

6cos⁡2x−cos⁡x−2=0 6\cos^2 x - \cos x - 2 = 0 6cos2x−cosx−2=0
[3]
b.

Hence, solve the equation

2sin⁡2kθ=4cos⁡2kθ−cos⁡kθ 2\sin^2 k\theta = 4\cos^2 k\theta - \cos k\theta 2sin2kθ=4cos2kθ−coskθ

where k k\,k is a positive integer, giving all values of θ \theta\,θ between 0∘ 0^\circ\,0∘ and 180∘ 180^\circ\,180∘ in terms of kkk.

[6]
Markscheme

Trigonometric Identities and Equations Questions

  1. A Level
  2. /Maths
  3. /Trigonometric Identities and Equations

322 exam-style questions on Edexcel A Level Maths Trigonometric Identities and Equations, covering 10.1 Angles in all four Quadrants, 10.2 Exact Values of Trigonometric Ratios, 10.3 Trigonometric Identities, 10.4 Solving Trigonometric Equations, 10.5 Harder Trigonometric Equations, and 10.6 Equations and Identities. Each one has a worked solution and a mark scheme showing where the marks go.

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