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)$\frac{3\pi}{8}$

Left side is  $ \sec^2 x $ and exploiting $2\sin x \cos x =\sin 2x  $ should ring a bell.
$$2\sin^2 2 x  = 1 \implies \sin 2x  = \pm \dfrac1{\sqrt2} $$ 
where $ x = \frac12 $ of $\pi/4 = \pi/8 $ in the first quadrant.
