$\begin{array}{l}
y = {\sin ^4}x + {\cos ^4}x\\
= {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\
= 1 - 2{\left( {\sin x\cos x} \right)^2}\\
= 1 - 2{\left( {\dfrac{1}{2}\sin 2x} \right)^2} = 1 - \dfrac{1}{2}{\sin ^2}2x\\
- 1 \le \sin 2x \le 1\\
\Rightarrow 0 \le {\sin ^2}2x \le 1\\
\Rightarrow 0 \le \dfrac{1}{2}{\sin ^2}2x \le \dfrac{1}{2}\\
\Rightarrow \dfrac{1}{2} \le y \le 1\\
\Rightarrow \left\{ \begin{array}{l}
\max y = 1 \Rightarrow {\sin ^2}2x = 0 \Rightarrow \sin 2x = 0\\
\min y = \dfrac{1}{2} \Rightarrow {\sin ^2}2x = 1 \Rightarrow \sin 2x = \pm 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\max y = 1 \Rightarrow 2x = k\pi \Rightarrow x = \dfrac{{k\pi }}{2}\\
\min y = \dfrac{1}{2} \Rightarrow 2x = \dfrac{\pi }{2} + k\pi \Rightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
