$\begin{array}{l}
{\sin ^2}3x = 4\cos 4x + 3\\
\dfrac{{1 - \cos 6x}}{2} = 4\cos 4x + 3\\
\Leftrightarrow 1 - \cos 6x = 8\cos 4x + 6\\
\Leftrightarrow 8\cos 4x + \cos 6x = - 5\\
\Leftrightarrow 8\left( {2{{\cos }^2}2x - 1} \right) + 4{\cos ^3}2x - 3\cos 2x = - 5\\
\Leftrightarrow 4{\cos ^3}2x + 16{\cos ^2}2x - 3\cos 2x - 3 = 0\\
\Leftrightarrow 4{\cos ^3}2x - 2{\cos ^2}2x + 18{\cos ^2}2x - 9\cos 2x + 6\cos 2x - 3 = 0\\
\Leftrightarrow 2{\cos ^2}2x\left( {2\cos 2x - 1} \right) + 9\cos 2x\left( {2\cos 2x - 1} \right) + 3\left( {2\cos 2x - 1} \right) = 0\\
\Leftrightarrow \left( {2\cos 2x - 1} \right)\left( {2{{\cos }^2}2x + 9\cos 2x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = \dfrac{1}{2}\\
\cos 2x = \dfrac{{ - 9 - \sqrt {57} }}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \pm \dfrac{\pi }{3} + k2\pi \\
2x = \pm \arccos \left( {\dfrac{{ - 9 +\sqrt {57} }}{4}} \right) + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{6} + k\pi \\
x = \dfrac{{ \pm \arccos \left( {\dfrac{{ - 9 +\sqrt {57} }}{4}} \right)}}{2} + k\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$
