Show that the equation
3sin2xtan2x=cos2x+2 3\sin 2x \tan 2x = \cos 2x + 2 3sin2xtan2x=cos2x+2Can be written in the form
4cos22x+2cos2x−3=0 4\cos^2 2x + 2\cos 2x - 3 = 0 4cos22x+2cos2x−3=0Find all values for x x\,x in the interval 0≤x<180∘0 \leq x < 180^\circ0≤x<180∘, for which
3sin2xtan2x=cos2x+2 3\sin 2x \tan 2x = \cos 2x + 2 3sin2xtan2x=cos2x+2Give your answers to two decimal places.