Let $\alpha ,\beta$ be any two positive values of x for which $2\cos x,\left|\cos x\right|$ and $1-3{\cos }^{2}x$ are in G.P. The minimum value of $\left|\alpha -\beta \right|$ is

${\cos }^{2}x=2\cos x\left(1-3{\cos }^{2}x\right)$

$\Rightarrow \cos x=0\text{ or }6{\cos }^{2}x+\cos x-2=0$

$\text{ Case (i): }\cos x=0\Rightarrow x=\frac{\pi }{2}$

$\text{ Case (ii): }6{\cos }^{2}x+4\cos x-3\cos x-2=0\Rightarrow \cos x=\frac{1}{2}\text{ or }\cos x=\frac{-2}{3}$

$\Rightarrow x=\frac{\pi }{2}orx=\pi -{\cos }^{-1}\left(\frac{2}{3}\right)$

If  $\alpha =\pi -{\cos }^{-1}\left(\frac{2}{3}\right)$ and  $\beta =\frac{\pi }{3}$,then $\alpha -\beta =\frac{2\pi }{3}-{\cos }^{-1}\left(\frac{2}{3}\right)$

$=\frac{2\pi }{3}-\frac{\pi }{3}=\frac{\pi }{3}$ and  $\left|\alpha -\beta \right|=\left|\frac{\pi }{2}-\frac{\pi }{3}\right|=\frac{\pi }{6}$

$\therefore$min $\left|\alpha -\beta \right|=\frac{\pi }{6}$