Show that the equation
3cosθ−2=5sinθtanθ 3\cos \theta - 2 = 5 \sin \theta \tan \theta 3cosθ−2=5sinθtanθcan be written in the form
8cos2θ−2cosθ−5=0 8\cos^2 \theta - 2\cos \theta - 5 = 0 8cos2θ−2cosθ−5=0Hence solve, for 0≤x<π0 \le x < \pi0≤x<π,
3cos2x−2=5sin2xtan2x 3\cos 2x - 2 = 5 \sin 2x \tan 2x 3cos2x−2=5sin2xtan2xgiving your answers, where appropriate, to 2 decimal places.
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.