Express \(6 \sin \theta + 8 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give your value for \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
Hence solve the equation \(6 \sin 2 x + 8 \cos 2 x = 7\), giving all solutions to the nearest \(0.1 ^ { \circ }\) in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\).
Prove the identity \(\frac { \sin 2 x } { 1 - \cos 2 x } = \frac { 1 } { \tan x }\).
Hence solve the equation
$$\frac { \sin 2 x } { 1 - \cos 2 x } = \tan x$$
giving all solutions in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\).