Show that the equation
2sinθcosθ3sinθ−2=3tanθ,sinθ≠23 \frac{2\sin\theta \cos\theta}{3\sin\theta - 2} = 3\tan\theta, \quad \sin\theta \neq \frac{2}{3} 3sinθ−22sinθcosθ=3tanθ,sinθ=32can be written in the form
2sin3θ+9sin2θ−8sinθ=0 2\sin^3\theta + 9\sin^2\theta - 8\sin\theta = 0 2sin3θ+9sin2θ−8sinθ=0Hence solve, for −π2<x<π2-\frac{\pi}{2} < x < \frac{\pi}{2}−2π<x<2π
2sinxcosx3sinx−2=3tanx \frac{2\sin x \cos x}{3\sin x - 2} = 3\tan x 3sinx−22sinxcosx=3tanxgiving your answers to 3 decimal places where appropriate.
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.