Given that \(\sqrt { 2 \sin ^ { 2 } \theta + \cos \theta } = 2 \cos \theta\), show that \(6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0\).
In this question you must show detailed reasoning.
Solve the equation
$$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
giving all values of \(\theta\) between \(0 ^ { \circ }\) and \(360 ^ { \circ }\) correct to 1 decimal place.
Explain why not all the solutions from part (ii) are solutions of the equation
$$\sqrt { 2 \sin ^ { 2 } \theta + \cos \theta } = 2 \cos \theta$$