Show that
cosθ(4tanθ+3tanθ)≡sinθ+3sinθ \cos \theta \left( 4 \tan \theta + \frac{3}{\tan \theta} \right) \equiv \sin \theta + \frac{3}{\sin \theta} cosθ(4tanθ+tanθ3)≡sinθ+sinθ3for θ≠nπ2\theta \neq \frac{n\pi}{2}θ=2nπ.
Hence solve, for 0<x<2π0 < x < 2\pi0<x<2π, the equation
cosx(4tanx+3tanx)=6sinx−1 \cos x \left( 4 \tan x + \frac{3}{\tan x} \right) = 6 \sin x - 1 cosx(4tanx+tanx3)=6sinx−1giving your answers to 3 significant figures.
137 exam-style questions on Edexcel A Level Maths Trigonometry and Modelling, covering 7.1 Addition Formulae, 7.2 Double Angle Formulae, 7.3 Solving Trigonometric Equations, 7.4 Simplifying a cos x +- b sin x, 7.5 Proving Trigonometric Identities, and 7.6 Modelling with Trigonometric Functions. Each one has a worked solution and a mark scheme showing where the marks go.