Show that the equation
2sin2x=4cos2x−cosx 2\sin^2 x = 4\cos^2 x - \cos x 2sin2x=4cos2x−cosxcan be expressed in the form
6cos2x−cosx−2=0 6\cos^2 x - \cos x - 2 = 0 6cos2x−cosx−2=0Hence, solve the equation
2sin22θ=4cos22θ−cos2θ 2\sin^2 2\theta = 4\cos^2 2\theta - \cos 2\theta 2sin22θ=4cos22θ−cos2θgiving all values of θ \theta\,θ between 0∘ 0^\circ\,0∘ and 180∘180^\circ180∘, correct to 1 decimal place.
