Show that the equation
2sin2x=7cosx+5 2\sin^2 x = 7\cos x + 5 2sin2x=7cosx+5can be written in the form
2cos2x+7cosx+3=0 2\cos^2 x + 7\cos x + 3 = 0 2cos2x+7cosx+3=0Hence solve, for 0≤x<360∘0 \leq x < 360^\circ0≤x<360∘, the equation
2sin2x=7cosx+5 2\sin^2 x = 7\cos x + 5 2sin2x=7cosx+511 exam-style questions on Edexcel A Level Maths 10.4 Solving Trigonometric Equations. Each one has a worked solution and a mark scheme showing where the marks go.