Explanation

(a)
$\because\ \cos{x = \sqrt{1 - \sin{2x}}}$
$\Rightarrow \ \cos{x = |\sin{x - \cos{x|}}}$
There are two cases arise.
Case I $\sin{x \leq \cos x}$
$\Rightarrow \cos{x = \cos{x - \sin x}}$
$\Rightarrow \sin{x = 0}$
where, $x \in \left\lbrack \ 0,\frac{\pi}{4} \right) \cup \left( \frac{5\pi}{4},\ 2\pi \right\rbrack$
$\Rightarrow x = 2\pi$, neglecting $x = \pi$
Case II $\sin{x > \cos x}$
$\Rightarrow \ \tan{x = 2}$
where, $x \in \left( \frac{\pi}{4},\frac{5\pi}{4} \right)$
$\because\ \tan{x = 2}$
$\Rightarrow \ x = \tan^{- 1}{(2)}$
Thus, the given equation has two solutions