$\begin{array}{l}
\sin x\cos 2x + {\cos ^2}x + {\cos ^2}x\left( {{{\tan }^2}x - 1} \right) + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{1}{2}\left( {\sin 3x - \sin x} \right) + {\cos ^2}x + {\cos ^2}x.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - {\cos ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{1}{2}\sin 3x - \dfrac{1}{2}\sin x + {\sin ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{1}{2}\left( {3\sin x - 4{{\sin }^3}x} \right) - \dfrac{1}{2}\sin x + {\sin ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow \dfrac{3}{2}\sin x - 2{\sin ^3}x - \dfrac{1}{2}\sin x + {\sin ^2}x + 2{\sin ^3}x = 0\\
\Leftrightarrow {\sin ^2}x + \dfrac{1}{2}\sin x = 0\\
\Leftrightarrow \sin x\left( {\sin x + \dfrac{1}{2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = - \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{ - \pi }}{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
2.\cos 3x + \cos 2x - \cos x - 1 = 0\\
\Leftrightarrow \left( {\cos 3x - \cos x} \right) + \left( {\cos 2x - 1} \right) = 0\\
\Leftrightarrow - 2\sin 2x\sin x + \left( {1 - 2{{\sin }^2}x - 1} \right) = 0\\
\Leftrightarrow - 2\sin 2x\sin x - 2{\sin ^2}x = 0\\
\Leftrightarrow - 2\sin x\left( {\sin 2x + \sin x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin 2x = - \sin x = \sin \left( { - x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2x = - x + k2\pi \\
2x = \pi + x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
3x = k2\pi \\
x = \pi + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
\end{array}$
