$\begin{array}{l}
{\sin ^4}x + {\cos ^4}x = - \dfrac{1}{2}{\cos ^2}2x\\
\Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x = - \dfrac{1}{2}{\cos ^2}2x\\
\Leftrightarrow 1 - 2{\left( {\dfrac{1}{2}\sin 2x} \right)^2} = - \dfrac{1}{2}{\cos ^2}2x\\
\Leftrightarrow 1 - \dfrac{1}{2}{\sin ^2}2x = - \dfrac{1}{2}{\cos ^2}2x\\
\Leftrightarrow \dfrac{1}{2}{\cos ^2}2x - \dfrac{1}{2}{\sin ^2}2x + 1 = 0\\
\Leftrightarrow \dfrac{1}{2}\left( {{{\cos }^2}2x - {{\sin }^2}2x} \right) = - 1\\
\Leftrightarrow \dfrac{1}{2}\cos 4x = - 1\\
\Leftrightarrow \cos 4x = - 2\left( 1 \right)\\
- 1 \le \cos 4x \le 1 \Rightarrow \left( 1 \right)VN\\
\Rightarrow S = \emptyset
\end{array}$
