[tex]4[sin(x)]^2 - 2 = 0\implies 4[sin(x)]^2=2\implies [sin(x)]^2=\cfrac{2}{4}\implies [sin(x)]^2=\cfrac{1}{2} \\\\\\ sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies sin^{-1}[sin(x)]=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)\implies x=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)[/tex]
[tex]x=sin^{-1}\left( \pm\cfrac{\sqrt{1}}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{1}{\sqrt{2}} \right)\implies x=sin^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill x=\cfrac{\pi }{4}~~,~~\cfrac{3\pi }{4}~~,~~\cfrac{5\pi }{4}~~,~~\cfrac{7\pi }{4}~\hfill[/tex]
Check the picture below.
