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Trigonometric Identities and Equations

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Question 70
a.

Show that the equation

2sin⁡2x=5cos⁡x−1 2\sin^2 x = 5\cos x - 1 2sin2x=5cosx−1

Can be written in the form

2cos⁡2x+5cos⁡x−3=0 2\cos^2 x + 5\cos x - 3 = 0 2cos2x+5cosx−3=0
[3]
b.

Hence solve, for 0≤x<360∘0 \leq x < 360^\circ0≤x<360∘, the equation,

2sin⁡2x=5cos⁡x−1 2\sin^2 x = 5\cos x - 1 2sin2x=5cosx−1
[5]
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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