Explanation

(b)
Given that, $\sin{x\sqrt{8\cos^{2}x} = 1}$
$\Rightarrow 2\sin{x|\cos{x| = \frac{1}{\sqrt{2}}}}$
If $\cos{x > 0}$, then $\sin{2x = \frac{1}{\sqrt{2}}}$
$\Rightarrow \ x = \frac{\pi}{8},\frac{3\pi}{8}$
And if $\cos{x < 0}$, then
$\sin{2x = - \frac{1}{\sqrt{2}}\ \Rightarrow \ x = \frac{5\pi}{8},\frac{7\pi}{8}\ }$
So, the required values of $x$ are $\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8}$, which form an AP with common difference $\frac{\pi}{4}$