Giải thích các bước giải:
Điều kiện:$ \left\{\begin{matrix}
sin^22x-4cos^2x\neq 0 & & \\
cos^2x\neq 0 & &
\end{matrix}\right.\Rightarrow x\neq \frac{\pi }{2}+k\pi$
$\frac{sin^2x-2}{sin^22x-4cos^2x}=tan^2x\\
\Leftrightarrow \frac{sin^2x-2}{4sin^2x.cos^2x-4cos^2x}=\frac{sin^2x}{cos^2x}\\
\Leftrightarrow \frac{sin^22x-2}{4(sin^2x-1)}=sin^2x\\
\Leftrightarrow sin^22x-2=4sin^2x(sin^2x-1)\\
\Leftrightarrow sin^22x-2=4sin^2x.(-cos^2x)\\
\Leftrightarrow 2sin^22x=2
\Leftrightarrow sin2x=\pm 1\Leftrightarrow x=\pm \frac{\pi}{4}+k\pi$
