$x_1,x_2$ are two positive values of $x$ for which $2 \cos x$, $|\cos x|$ and $(3\sin^2x-2)$ are in geometric sequence. One possible value of $|x_1-x_2|$ can be equal to
a) $\frac{2\pi}{3}$ b) $\frac{\pi}{3}$ c) $2\cos^{-1}(\frac{2}{3})$ d) $\cos^{-1}(\frac{2}{3})$
Since $2 \cos x$, $|\cos x|$ and $(3\sin^2x-2)$ are in geometric progression, therefore,$$|\cos x|^2 = 2\cos x (3\sin^2x-2) .$$
The right hand side is:$$2\cos x(3(1-\cos^2x)-2) = 6\cos x - 6\cos^3x-4\cos x .$$
So \begin{gather*} \cos x = 6-6\cos^2x-4 = 2-6\cos^2x\\ \Rightarrow \quad 6\cos^2x-\cos x-2=0 \end{gather*}
Therefore, \begin{gather*} \cos x = \frac{1 \pm \sqrt{1+24}}{12}\\ \Rightarrow \quad \cos x = \frac{1}{2}, \; \frac{-1}{3} \end{gather*}
Please guide further....
I think you were on the right track, there was just a small mistake. I understand the problem as follows:
$$2\cos(x)=a,\quad |\cos(x)|=ar,\quad 3\sin^2(x)-2=ar^2$$
We get $$r=\frac{|\cos(x)|}{2\cos(x)}=\pm \frac{1}{2}$$ and consequently $$3\sin^2(x)-2 = 1-3\cos^2(x) = \frac{a}{4} = \frac{1}{2}\cos(x)$$
which gives
$$6\cos^2(x) + \cos(x) - 2 = 0$$
with the solutions
$$\cos(x) = \{\frac{1}{2}, -\frac{2}{3} \}$$