Explanation

(b)
$\because\ \sec{x - 1 = \left( \sqrt{2} - 1 \right)\tan x}$
$\Rightarrow \ 1 - \cos{x = \left( \sqrt{2} - 1 \right)\sin x}$
$\Rightarrow \ \sin\frac{x}{2}\left\{ \sin{\frac{x}{2} - \left( \sqrt{2} - 1 \right)\cos\frac{x}{2}} \right\} = 0$
$\Rightarrow \sin{\frac{x}{2} = 0\ or\ \tan{\frac{x}{2} = \sqrt{2} - 1 = \tan\frac{\pi}{8}}}$
$\Rightarrow \frac{x}{2} = n\pi\ or\ \frac{x}{2} = n\pi + \frac{\pi}{8}$
$\therefore\ x = 2n\pi,\ 2n\pi + \frac{\pi}{4}$