Example 3. Solve the system of equations
$$
\left\{\begin{array}{l}
x \sin a+y \sin 2 a+z \sin 3 a=\sin 4 a, \\
x \sin b+y \sin 2 b+z \sin 3 b=\sin 4 b, \\
x \sin c+y \sin 2 c+z \sin 3 c=\sin 4 c .
\end{array}\right.
$$
By applying the double, triple, and quadruple angle formulas for $\sin \alpha$, equation (1) can be transformed into
$$
\begin{array}{l}
8 \cos ^{3} a-4 z \cos ^{2} a-(2 y+4) \cos a+z-x \\
=0 .
\end{array}
$$

Let $t=\cos a$, and divide both sides of the equation by 8, we get the equation in terms of $t$:
$$
t^{3}-\frac{z}{2} t^{2}+\frac{y+2}{4} t+\frac{z-x}{8}=0 .
$$

This shows that $\cos a$ is a root of equation (1'). Considering the symmetry of the original system of equations, we know that $\cos b$ and $\cos c$ are also roots of equation (1'). Therefore, by the generalized Vieta's formulas, we have
$$
\left\{\begin{array}{l} 
\frac{z}{2}=\cos a+\cos b+\cos c, \\
-\frac{y+2}{4}= \cos a \cos b+\cos a \cos c \\
+\cos b \cos c, \\
-\frac{z-x}{8}= \cos a \cos b \cos c.
\end{array}\right.
$$

Thus, the solution to the original system of equations is:
$$
\left\{\begin{aligned}
x= & 2(\cos a+\cos b+\cos c) \\
& +8 \cos a \cos b \cos c, \\
y= & -2-4(\cos a \cos b+\cos a \cos c \\
& +\cos b \cos c), \\
z= & 2(\cos a+\cos b+\cos c) .
\end{aligned}\right.
$$