<s>[INST] 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)$\frac{3\pi}{8}$
 [/INST] Use the identity $$1+\tan^2(x) = \sec^2(x)$$ Then multiply your equation through by $\cos^2(x)$ to get $$\begin{align*}1 = 8\sin^2(x)\cos^2(x) \\ = 2(2\sin(x)\cos(x))^2 \\ = 2(\sin(2x))^2 \\ \implies \frac{1}{2} = (\sin(2x))^2 \\ \implies \pm \frac{\sqrt{2}}{2} = \sin(2x)\end{align*}$$ Can you take it from here?
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