Inverse trigonometric question including geometric sequence

$x_1,x_2$ are two positive values of x for which 2 cosx, $|cosx|$ and $(3sin^2x-2)$ 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) $2cos^{-1}(\frac{2}{3})$ d) $cos^{-1}(\frac{2}{3})$

Since 2 cosx, $|cosx|$ and $(3sin^2x-2)$ are in G.P therefore, $|cosx|^2 = 2cosx (3sin^2x-2)$ R.H.S. = $2cosx(3(1-cos^2x)-2)$ = $6cosx - 6cos^3x-4cosx$

$\Rightarrow cosx = 6-6cos^2x-4 = 2-6cos^2x$

$\Rightarrow 6cos^2x-cosx-2=0$

Therefore, cosx = $\frac{1 \pm \sqrt{1+24}}{12}$ $\Rightarrow cosx = \frac{1}{2}; \frac{-1}{3}$

-

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

$$\cos(x) = \{\frac{1}{2}, -\frac{2}{3} \}$$

-
But answer options are different could u please elaborate further... – sultan Apr 15 '13 at 11:54
So you still have to find $|x_1-x_2|$ when $\cos x_1=1/2$ and $\cos x_2=-2/3$, but presumably you know how to find the values of $x_1$ and $x_2$ that give you those cosines. – Gerry Myerson Apr 15 '13 at 11:56