Question 9 - A-Level Maths

Prove that $$\cos ^ { 2 } \left( \theta + 45 ^ { \circ } \right) - \frac { 1 } { 2 } ( \cos 2 \theta - \sin 2 \theta ) \equiv \sin ^ { 2 } \theta .$$
Hence solve the equation $$6 \cos ^ { 2 } \left( \frac { 1 } { 2 } \theta + 45 ^ { \circ } \right) - 3 ( \cos \theta - \sin \theta ) = 2$$ for $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$.
It is given that there are two values of $\theta$, where $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$, satisfying the equation $$6 \cos ^ { 2 } \left( \frac { 1 } { 3 } \theta + 45 ^ { \circ } \right) - 3 \left( \cos \frac { 2 } { 3 } \theta - \sin \frac { 2 } { 3 } \theta \right) = k ,$$ where $k$ is a constant. Find the set of possible values of $k$.