Example 4. Solve the trigonometric equation: $\sin 3 x=\sin 2 x$.
Given $z=\cos x+i \sin x$.
$$
\frac{z^{3}-1}{2 i z^{3}}=\frac{z^{2}-1}{2 i z}
$$

Expanding and rearranging, we get $z^{\theta}-z^{4}+z^{2}-1=0$.
Solving the equation, we have
$$
\begin{array}{l}
\left(z^{2}-1\right)\left(z^{4}+1\right)=0 . \\
\text { When } z^{2}-1=0, \\
z^{2}=1, \text { i.e., } \cos 2 x+i \sin 2 x=1 .
\end{array}
$$

Comparing the real part coefficients, we get $\cos 2 x=1$, thus $2 x=2 k \pi$,
$$
\begin{aligned}
x=k \pi(k \in \mathbb{Z}) \\
\text { When } z^{4}+1=0, \text { then } z^{4}=-1,
\end{aligned}
$$

Thus $\cos 4 x+i \sin 4 x=-1$.
Comparing the real part coefficients, we get $\cos 4 x=-1$, thus
$$
\begin{aligned}
4 x & =2 k \pi-\pi \\
x & =\frac{1}{4}(2 k-1) \pi(k \in \mathbb{Z}) .
\end{aligned}
$$

Therefore, the solutions are
$$
\begin{array}{l}
\{x \mid x=k \pi, k \in \mathbb{Z}\} \\
\cup\left\{x \left\lvert\, x=\frac{1}{4}(2 k-1) \pi\right., k \in \mathbb{Z}\right\} \cdot
\end{array}
$$