Show that 6cos2θ+7sinθ−81−2sinθ≡3sinθ−2\displaystyle \frac{6\cos^2 \theta + 7\sin \theta - 8}{1 - 2\sin \theta} \equiv 3\sin \theta - 21−2sinθ6cos2θ+7sinθ−8≡3sinθ−2
Hence solve, for 0≤θ<360∘0 \leq \theta < 360^\circ0≤θ<360∘, the equation, 6cos2θ+7sinθ−81−2sinθ=2cosθ−2\displaystyle \frac{6\cos^2 \theta + 7\sin \theta - 8}{1 - 2\sin \theta} = 2\cos \theta - 21−2sinθ6cos2θ+7sinθ−8=2cosθ−2
41 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.