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Trigonometric Identities and Equations

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Question 56
a.

Show that the equation

4sin⁡2x=16cos⁡2x−11cos⁡x 4\sin^2 x = 16\cos^2 x - 11\cos x 4sin2x=16cos2x−11cosx

can be expressed in the form

20cos⁡2x−11cos⁡x−4=0 20\cos^2 x - 11\cos x - 4 = 0 20cos2x−11cosx−4=0
[3]
b.

Hence, solve the equation

4sin⁡22θ=16cos⁡22θ−11cos⁡2θ 4\sin^2 2\theta = 16\cos^2 2\theta - 11\cos 2\theta 4sin22θ=16cos22θ−11cos2θ

giving all values of θ \theta\,θ between 0∘ 0^\circ\,0∘ and 180∘180^\circ180∘, correct to 1 decimal place.

[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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