Show that the equation
2sin2x=4cos2x−cosx 2\sin^2 x = 4\cos^2 x - \cos x 2sin2x=4cos2x−cosxcan be expressed in the form
6cos2x−cosx−2=0 6\cos^2 x - \cos x - 2 = 0 6cos2x−cosx−2=0Hence, solve the equation
2sin22θ=4cos22θ−cos2θ 2\sin^2 2\theta = 4\cos^2 2\theta - \cos 2\theta 2sin22θ=4cos22θ−cos2θgiving all values of θ \theta\,θ between 0∘ 0^\circ\,0∘ and 180∘180^\circ180∘, correct to 1 decimal place.
Practise AQA A Level Maths 1.8 E: Trigonometry with exam-style questions for A Level Maths. 239 questions covering 1.8.1 Trigonometric definitions, rules and radians, 1.8.2 Small angle approximations (A-level only), 1.8.3 Trigonometric functions and exact values, 1.8.4 Reciprocal and inverse trigonometric functions (A-level only), 1.8.5 Trigonometric identities, 1.8.6 Compound and double angle formulae (A-level only), 1.8.7 Solving trigonometric equations, 1.8.8 Proofs with trigonometric identities, and 1.8.9 Trigonometry in context, matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.