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+5