Express \(\cos \theta + ( \sqrt { } 3 ) \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving the exact value of \(\alpha\).
Hence show that one solution of the equation
$$\cos \theta + ( \sqrt { } 3 ) \sin \theta = \sqrt { } 2$$
is \(\theta = \frac { 7 } { 12 } \pi\), and find the other solution in the interval \(0 < \theta < 2 \pi\).
By sketching a suitable pair of graphs, for \(x < 0\), show that exactly one root of the equation \(x ^ { 2 } = 2 ^ { x }\) is negative.
Verify by calculation that this root lies between - 1.0 and - 0.5 .
Use the iterative formula
$$x _ { n + 1 } = - \sqrt { } \left( 2 ^ { x _ { n } } \right)$$
to determine this root correct to 2 significant figures, showing the result of each iteration.