Show that the equation 3cosθ−2=5sinθtanθ3\cos \theta - 2 = 5 \sin \theta \tan \theta3cosθ−2=5sinθtanθ can be written in the form 8cos2θ−2cosθ−5=08\cos^2 \theta - 2\cos \theta - 5 = 08cos2θ−2cosθ−5=0
Hence solve, for 0≤x<π0 \le x < \pi0≤x<π, 3cos2x−2=5sin2xtan2x3\cos 2x - 2 = 5 \sin 2x \tan 2x3cos2x−2=5sin2xtan2x giving your answers, where appropriate, to 2 decimal places.