Show that the equation 1+cosx=3sin2xcosx1 + \cos x = \dfrac{3\sin^2 x}{\cos x}1+cosx=cosx3sin2x can be written in the form 4cos2x+cosx−3=04\cos^2 x + \cos x - 3 = 04cos2x+cosx−3=0
Hence solve, for 0≤x<360∘0 \leq x < 360^\circ0≤x<360∘, the equation, 1+cosx=3sin2xcosx1 + \cos x = \dfrac{3\sin^2 x}{\cos x}1+cosx=cosx3sin2x. Give your answers to one decimal place where appropriate.
322 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.