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)$\frac{3\pi}{8}$
 A: Recall $1+\tan^2(x) = \sec^2(x)$ and $\sin(2x) = 2\sin(x)\cos(x)$. Hence,
$$1+\tan^2(x) = 8 \sin^2(x) \implies \sec^2(x) = 8\sin^2(x) \implies 8\sin^2(x)\cos^2(x) = 1$$
This gives us
$$2\sin^2(2x) = 1 \implies \sin(2x) = \pm \dfrac1{\sqrt2} \implies 2x = \dfrac{n\pi}2 + \dfrac{\pi}4 \implies x = \dfrac{n\pi}4 + \dfrac{\pi}8$$
A: 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 $x$.
A: 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.
A: 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?
A: 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
A: Since this is a multiple-choice question, you can also check each of the available choices and see which ones work.  Notice that several of the choices satisfy the equation but are outside the interval $ [ \pi/6 , \pi/2]$, so can be excluded.  In fact three of the six choices can be excluded based on that consideration alone.  For the remaining three, just plug them in and see if they are correct.
