- (a) Show that
$$\frac { 10 \sin ^ { 2 } \theta - 7 \cos \theta + 2 } { 3 + 2 \cos \theta } \equiv 4 - 5 \cos \theta$$
(b) Hence, or otherwise, solve, for \(0 \leqslant x < 360 ^ { \circ }\), the equation
$$\frac { 10 \sin ^ { 2 } x - 7 \cos x + 2 } { 3 + 2 \cos x } = 4 + 3 \sin x$$