Explanation

(d)
$\because\ 2\cos{x,\ |\cos{x|\ \ ,\ 1 - 3\cos^{2}x}}$ are in GP
$\therefore\ \cos^{2}{x = 2\cos{x \bullet (1 - 3\cos^{2}{x)}}}$
$\Rightarrow \ 6\cos^{3}{x + \cos^{2}{x - 2\cos{x = 0}}}$
$\therefore\ \cos{x = 0,\frac{1}{2},\ - \frac{2}{3}}$
$\therefore\ x = \frac{\pi}{2},\frac{\pi}{3},\cos^{- 1}\left( \ - \frac{2}{3} \right)$ ($\because\ \alpha,\ \beta$ are positive)
If $\alpha = \frac{\pi}{2},\ \beta = \frac{\pi}{3}$
Then, $|\alpha - \beta| = \frac{\pi}{6}$