Given \(\sin x-\cos x=0\)

\[

\begin{aligned}

& \Rightarrow \sin x=\cos x \\

& \Rightarrow \tan x=1 \\

& \therefore \tan x=\tan \frac{\pi}{4} \quad \text { as } \tan 45^{\circ}=1 \\

& \Rightarrow x=\frac{\pi}{4}

\end{aligned}

\]

\[

\begin{aligned}

\therefore \sin ^4 x+\cos ^2 x & =\sin ^4 \frac{\pi}{4}+\cos ^4 \frac{\pi}{4} \\

& =\left(\frac{1}{\sqrt{2}}\right)^4+\left(\frac{1}{\sqrt{2}}\right)^4 \\

& =\frac{1}{4}+\frac{1}{4}

\end{aligned}

\]

\[\sin ^4 x+\cos ^2 x=\frac{1}{2}\]