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.
41 exam-style questions on Edexcel AS 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.