Giải phương trình câu này giúp mình với

$\begin{array}{l}
\sin x\cos 2x = \sin 2x\cos 3x\\
 \Leftrightarrow \sin x\cos 2x = 2\sin x\cos x\cos 3x\\
 \Leftrightarrow \sin x\left( {\cos 2x - 2\cos x\cos 3x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos 2x = 2\cos x\cos 3x
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\cos 2x = 2\cos x\left( {4{{\cos }^3}x - 3\cos x} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2{\cos ^2}x - 1 = 8{\cos ^4}x - 6{\cos ^2}x
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
8{\cos ^4}x - 8{\cos ^2}x + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
8{t^2} - 8t + 1 = 0\left( {t = {{\cos }^2}x,0 \le t \le 1} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
t = \dfrac{{2 + \sqrt 2 }}{4}\\
t = \dfrac{{2 - \sqrt 2 }}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
{\cos ^2}x = \dfrac{{2 + \sqrt 2 }}{4}\\
{\cos ^2}x = \dfrac{{2 - \sqrt 2 }}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\dfrac{{1 + \cos 2x}}{2} = \dfrac{{2 + \sqrt 2 }}{4}\\
\dfrac{{1 + \cos 2x}}{2} = \dfrac{{2 - \sqrt 2 }}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2 + 2\cos 2x = 2 + \sqrt 2 \\
2 + 2\cos 2x = 2 - \sqrt 2 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
\cos 2x = \dfrac{{\sqrt 2 }}{2}\\
\cos 2x =  - \dfrac{{\sqrt 2 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
2x =  \pm \dfrac{\pi }{4} + k2\pi \\
2x =  \pm \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x =  \pm \dfrac{\pi }{8} + k\pi \\
x =  \pm \dfrac{{3\pi }}{8} + k\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{\pi }{8} + k\dfrac{\pi }{4}
\end{array} \right.\left( {k \in Z} \right)
\end{array}$