Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.

$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$

I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful.
I need to find the angle "x" which is:
a)$\frac{\pi}{8}\quad$    b)$\frac{\pi}{6}\quad$  c)$\frac{\pi}{4}\quad$ d)$\frac{5\pi}{6}\quad$ e)$\frac{3\pi}{4}\quad$ f)$\frac{3\pi}{8}$

Here's another approach.  Use the facts that $\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}$ and $\cos^2(x) = 1 - \sin^2(x)$ to rewrite the equation so that the only trigonometric expression in it is $\sin^2(x)$.  Then let $u = \sin^2(x)$.  Now your equation should read:
$$1 + \frac{u}{1-u} = 8u$$
Now solve for $u$, then back-substitute in $u = \sin^2(x)$ to your solution(s) for $u$ and solve for $x$.
