Explanation

(a)
Let $\sqrt{3} + 1 = r\cos{\alpha\ and\ \sqrt{3} - 1 = r\sin\alpha}$, then
$r = \sqrt{\left( \sqrt{3} + 1 \right)^{2} + \left( \sqrt{3} - 1 \right)^{2}\ }$
$= \sqrt{3 + 1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3}} = 2\sqrt{2}$
and $\tan{\alpha = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{1 - \left( \ \frac{1}{\sqrt{3}} \right)}{1 + \left( \ \frac{1}{\sqrt{3}} \right)} = \tan\left( \frac{\pi}{4} - \frac{\pi}{6} \right)}$
$\Rightarrow \ \alpha = \frac{\pi}{12}$
The given equation reduces to
$2\sqrt{2}\cos{(\theta - \alpha) = 2}$
$\Rightarrow \ \cos{\left( \ \theta - \frac{\pi}{12} \right) = \cos\frac{\pi}{4}}$
$\Rightarrow \ \theta - \frac{\pi}{12} = 2n\pi \pm \frac{\pi}{4}$
$\Rightarrow \ \theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$