$\begin{array}{l}
\sin 3x = \sqrt 2 \cos 4x - \cos 3x\\
\Leftrightarrow \sin 3x + \cos 3x = \sqrt 2 \cos 4x\\
\Leftrightarrow \sqrt 2 \cos \left( {3x - \dfrac{\pi }{4}} \right) = \sqrt 2 \cos 4x\\
\Leftrightarrow \left[ \begin{array}{l}
3x - \dfrac{\pi }{4} = 4x + k2\pi \\
3x - \dfrac{\pi }{4} = \pi - 4x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{4} - k2\pi \\
x = \dfrac{{5\pi }}{{28}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
b)\sin 2x + {\sin ^2}x = \dfrac{1}{2}\\
\Leftrightarrow \sin 2x + \dfrac{{1 - \cos 2x}}{2} = \dfrac{1}{2}\\
\Leftrightarrow 2\sin 2x - \cos 2x = 0\\
\Leftrightarrow \sqrt 5 \left( {\dfrac{2}{{\sqrt 5 }}\sin 2x - \dfrac{1}{{\sqrt 5 }}\cos 2x} \right) = 0\\
\Leftrightarrow \sin \left( {2x - \alpha } \right) = 0\left( {\alpha = \arccos \dfrac{2}{{\sqrt 5 }}} \right)\\
\Leftrightarrow 2x - \alpha = k\pi \Leftrightarrow x = \dfrac{\alpha }{2} + \dfrac{{k\pi }}{2}
\end{array}$
