bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,
[tex]\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2}
\\\\\\
sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm\sqrt{1}
\\\\\\
sin(2x)=\pm 1\implies sin^{-1}[sin(2x)]=sin^{-1}(\pm 1)[/tex]
[tex]\bf \measuredangle 2x=sin^{-1}(\pm 1)\implies \measuredangle 2x=
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}\\\\
-------------------------------\\\\
\measuredangle 2x=\cfrac{\pi }{2}\implies \measuredangle x=\cfrac{\pi }{4}\qquad \qquad \qquad \qquad \measuredangle 2x=\cfrac{3\pi }{2}\implies \measuredangle x=\cfrac{3\pi }{4}[/tex]
