$\begin{array}{l}
{\sin ^4}x + \cos {\left( {x + \dfrac{\pi }{4}} \right)^4} = \dfrac{1}{4}\\
\Leftrightarrow {\left( {\dfrac{{1 - \cos 2x}}{2}} \right)^2} + {\left[ {\dfrac{{1 + \cos \left( {2x + \dfrac{\pi }{2}} \right)}}{2}} \right]^2} = \dfrac{1}{4}\\
\Leftrightarrow \dfrac{{{{\left( {1 - \cos 2x} \right)}^2}}}{4} + \dfrac{{{{\left[ {1 + \cos \left( {2x + \dfrac{\pi }{2}} \right)} \right]}^2}}}{4} = \dfrac{1}{4}\\
\Leftrightarrow {\left( {1 - \cos 2x} \right)^2} + {\left( {1 - \sin 2x} \right)^2} = 1\\
\Leftrightarrow {\cos ^2}2x - 2\cos 2x + 1 + 1 + {\sin ^2}2x - 2\sin 2x = 1\\
\Leftrightarrow 3 - 2\left( {\sin 2x + \cos 2x} \right) = 1\\
\Leftrightarrow \sin 2x + \cos 2x = 1\\
\Leftrightarrow \sqrt 2 \sin \left( {2x + \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow \sin \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
2x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
