$\begin{array}{l}
\sin 2x - \cos 2x = \sqrt 2 \\
\Leftrightarrow \sqrt 2 \left( {\dfrac{{\sqrt 2 }}{2}\sin 2x - \dfrac{{\sqrt 2 }}{2}\cos 2x} \right) = \sqrt 2 \\
\Leftrightarrow \cos \dfrac{\pi }{4}\sin 2x - \sin \dfrac{\pi }{4}\cos 2x = 1\\
\Leftrightarrow \sin \left( {2x - \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow 2x - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow x = \dfrac{{3\pi }}{8} + k\pi \left( {k \in \mathbb{Z}} \right)
\end{array}$
