Example 4. Solve the equation $\sin x + \sin 2x + \sin 3x = 0$.
Solution 1 $\sin x+\sin 2 x+\sin 3 x$
$$
\begin{array}{l}
=(\sin x+\sin 3 x)+\sin 2 x \\
=2 \sin 2 x \cdot \cos x+\sin 2 x=\sin 2 x(2 \cos x+1) .
\end{array}
$$

Thus, the original equation becomes
$$
\sin 2 x \cdot(2 \cos x+1)=0 \text {. }
$$

Solving it, we get: $\sin 2 x=0, \cos x=-\frac{1}{2}$.
$\therefore$ The solution set is $\left\{x \left\lvert\, x=\frac{n \pi}{2}\right., \mathrm{n} \in \mathrm{J}\right\}$
$$
U\left\{x \left\lvert\, x=2 n \pi \pm \frac{2 \pi}{3}\right., \mathrm{n} \in \mathrm{J}\right\} .
$$

Solution 2 Since $\sin x+\sin 2 x+\sin 3 x$
$$
\begin{array}{l}
=2 \sin \frac{3 x}{2} \cdot \cos \frac{x}{2}+2 \sin \frac{3 x}{2} \cdot \cos \frac{3 x}{2} \\
=2 \sin \frac{3 x}{2}-\left(\cos \frac{x}{2}+\cos \frac{3 x}{2}\right),
\end{array}
$$

Thus, the original equation is $2 \sin \frac{3 x}{2}\left(\cos \frac{x}{2}+\cos \frac{3 x}{2}\right)=0$, i.e., $4 \sin \frac{3 x}{2} \cdot \cos \frac{x}{2} \cdot \cos x=0$.
Solving it, we get $\sin \frac{3 x}{2}=0, \quad \cos \frac{x}{2}=0$, $\cos x=0$.
Therefore, the solution set is $\left\{x \left\lvert\, x=-\frac{2 n \pi}{3}\right., n \in J\right\} U$
$$
\begin{array}{l}
\left\{x \left\lvert\, x=2 n \pi \pm \frac{\pi}{2}\right., n \in J\right\} U\{x \mid x=4 n \pi \pm x, \\
n \in J\} .
\end{array}
$$

As shown in Figure 6, drawing the terminal sides of the angles in (1) and (2) on the unit circle, we can see that the terminal sides of the angles $x$ in (1) and (2) are completely overlapping, and there are no exceptions. This shows that (1) and (2) are two different representations of the solution set of the equation, and they are all solutions to the original equation.