If \(\cos ^2\left[\frac{\pi}{4}\left(\sin x+\sqrt{2} \cos ^2 x\right)\right]-\tan ^2\left(x+\frac{\pi}{4} \tan ^2 x\right)=1\) then what is the value of \(\sin x\) ?
\(\cos ^2\left[\frac{\pi}{4}\left(\sin x+\sqrt{2} \cos ^2 x\right)\right]-\tan ^2\left(x+\frac{\pi}{4} \tan ^2 x\right)=1\)
\(-\tan ^2\left(x+\frac{\pi}{4} \tan ^2 x\right)=1-\cos ^2\left(\frac{\pi}{4}\left(\sin x+\sqrt{2} \cos ^2 x\right)\right)\)
\(\sin ^2\left(\frac{\pi}{4}\left(\sin x+\sqrt{2} \cos ^2 x\right)\right)+\tan ^2\left(x+\frac{\pi}{4} \tan ^2 x\right)=0 \)
\(\sin \left(\frac{\pi}{4}\left(\sin x+\sqrt{2} \cos ^2 x\right)\right)=0 \)
\(\frac{\pi}{4}\left(\sin x+\sqrt{2} \cos ^2 x\right)=n \pi\)
\(\sin x+\sqrt{2}\left(1-\sin ^2 x\right)=4 n\)
\(\sin x+\sqrt{2}-\sqrt{2} \sin ^2 x=0\)
\(\sqrt{2} \sin^2 x-\sin x-\sqrt{2}=0\)
\(\sqrt{2} \sin^2 x-2 \sin x+\sin x-\sqrt{2}=0\)
\( \sqrt{2} \sin x(\sin x-\sqrt{2})+1(\sin x-\sqrt{2})=0 \)
\((\sqrt{2} \sin x+1)(\sin x-\sqrt{2})=0\)
\(\sin x=-\frac{1}{\sqrt{2}}, \sqrt{2}\)
