Giải cos3x+cos2x=0 Giúp mk gấp vs ạ

$\begin{array}{l}
\cos 3x + \cos 2x = 0\\
 \Leftrightarrow \cos 3x =  - \cos 2x\\
 \Leftrightarrow \cos 3x = \cos \left( {\pi  - 2x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = \pi  - 2x + k2\pi \\
3x = 2x - \pi  + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
5x = \pi  + k2\pi \\
x =  - \pi  + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{5} + \dfrac{{k2\pi }}{5}\\
x =  - \pi  + k2\pi 
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$