- (a) Show that the equation
$$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$
can be written in the form
$$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
(b) Hence solve, for \(0 \leqslant x < 90 ^ { \circ }\)
$$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$
giving your answers, where appropriate, to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.)