<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] If $\sin^2y=\sin^2A\iff\cos2y=1-2\sin^2y=\cdots=\cos2A$
$\iff2y=2m\pi\pm2A\iff y=m\pi\pm A$ where $m$ is any integer
We have $\sin^2(2x)=\dfrac12=\left(\sin\dfrac\pi4\right)^2$
$\implies2x=r\pi\pm\dfrac\pi4=\dfrac\pi4\left(4r\pm1\right)$ where $r$ is any integer
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