Can I solve x on this way? or is there a mistake I made or is the question wrong? $$\cos3x\sin2x=\cos3x$$
Can I just pull $\cos3x$ out of both sides and continue with $\sin2x=0$ which leaves us with $2x=0$?
We have
$$
\cos(3x) \sin(2x) = \cos(3x) \iff \\
\cos(3x) (\sin(2x) - 1) = 0 \iff \\
\cos(3x) = 0 \vee (\sin(2x) - 1) = 0 \iff \\
\cos(3x) = 0 \vee \sin(2x) = 1 \iff \\
3x = \frac{\pi}{2} + k \pi \vee 2x = \frac{\pi}{2} + m 2\pi \iff \\
x = \frac{\pi}{6} + k \frac{\pi} 3 \vee x = \frac{\pi}{4} + m \pi
$$
for arbitrary $k, m \in \mathbb{Z}$.
$\cos 3x \sin 2x=\cos 3x\implies \cos 3x=0$ or $\sin 2x=1$

The rest you can solve.$0/0$ is undefined.Keep that in mind.

If any bound on $x$ is given, then answer it appropriately.