- (a) Use the identity \(\cos ( A + B ) = \cos A \cos B - \sin A \sin B\), to show that
$$\cos 2 A = 1 - 2 \sin ^ { 2 } A$$
The curves \(C _ { 1 }\) and \(C _ { 2 }\) have equations
$$\begin{aligned}
& C _ { 1 } : \quad y = 3 \sin 2 x
& C _ { 2 } : \quad y = 4 \sin ^ { 2 } x - 2 \cos 2 x
\end{aligned}$$
(b) Show that the \(x\)-coordinates of the points where \(C _ { 1 }\) and \(C _ { 2 }\) intersect satisfy the equation
$$4 \cos 2 x + 3 \sin 2 x = 2$$
(c) Express \(4 \cos 2 x + 3 \sin 2 x\) in the form \(R \cos ( 2 x - \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) to 2 decimal places.
(d) Hence find, for \(0 \leqslant x < 180 ^ { \circ }\), all the solutions of
$$4 \cos 2 x + 3 \sin 2 x = 2$$
giving your answers to 1 decimal place.