4・153 Solve the system of equations
$$
\left\{\begin{array}{l}
\sin x + 2 \sin (x + y + z) = 0 \\
\sin y + 3 \sin (x + y + z) = 0 \\
\sin z + 4 \sin (x + y + z) = 0 .
\end{array}\right.
$$
[Solution] (1) + (2) - (3) gives
$$
\sin x+\sin y-\sin z+\sin (x+y+z)=0,
$$

Using sum-to-product identities, we get
$$
\begin{array}{l}
2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}+2 \cos \left(\frac{x+y}{2}+z\right) \sin \frac{x+y}{2}=0, \\
\sin \frac{x+y}{2} \cos \frac{x+z}{2} \cos \frac{y+z}{2}=0 .
\end{array}
$$

Thus, we have $\sin \frac{x+y}{2}=0$ or $\cos \frac{x+z}{2}=0$ or $\cos \frac{y+z}{2}=0$.
Case 1 If $\sin \frac{x+y}{2}=0$, then
$$
\frac{x+y}{2}=k \pi \text { ( } k \text { is any integer), }
$$

i.e., $x+y=2 k \pi$.
Substituting (4) into the original system of equations, we get
$$
\left\{\begin{array}{l}
\sin x+2 \sin z=0, \\
\sin y+3 \sin z=0, \\
\sin z+4 \sin z=0 .
\end{array}\right.
$$

From (7), we get $\sin z=0$,
thus $z=m \pi(m$ is any integer $)$.
Substituting into (5) and (6), we get $\left\{\begin{array}{l}\sin x=0, \\ \sin y=0 .\end{array}\right.$
Therefore, $x=n \pi, y=l \pi(n, l$ are any integers $)$.
Upon verification, the original system of equations has the solution
$$
\left\{\begin{array}{l}
x=n \pi, \\
y=l \pi,(m, n, l \text { are any integers }) \\
z=m \pi .
\end{array}\right.
$$

Case 2 If $\cos \frac{x+z}{2}=0$, then similarly we get the solution (8).
Case 3 If $\cos \frac{y+z}{2}=0$, then similarly we get the solution (8).
In summary, the solution to the original system of equations is (8).