9. Solve the system of equations $\left\{\begin{array}{l}\sin x+2 \sin (x+y+z)=0 \\ \sin y+3 \sin (x+y+z)=0 \\ \sin z+4 \sin (x+y+z)=0\end{array}\right.$
9. (1) + (2) - (3) yields $\sin x + \sin y + \sin (x + y + z) - \sin z = 0, 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2} + 2 \cos \frac{x + y + z}{2} \sin \frac{x + y}{2} = 0, 2 \sin \frac{x + y}{2} \left( \cos \frac{x - y}{2} + \cos \frac{x + y + 2z}{2} \right) = 0, 4 \sin \frac{x + y}{2} \cos \frac{x + z}{2} \cos \frac{y + z}{2} = 0, \sin \frac{x + y}{2} \cos \frac{x + z}{2} \cos \frac{y + z}{2} = 0, \sin \frac{x + y}{2} = 0$ or $\cos \frac{x + z}{2} = 0$ or $\cos \frac{y + z}{2} = 0$
(1) If $\sin \frac{x + y}{2} = 0$, then $\frac{x + y}{2} = k \pi \Rightarrow x + y = 2k \pi$, substituting into (3) gives $\sin z = 0 \Rightarrow z = k_1 \pi$, substituting into (1) and (2) respectively gives $x = k_2 \pi, y = k_3 \pi$;
(2) In the cases where $\cos \frac{x + z}{2} = 0$ and $\cos \frac{y + z}{2} = 0$, the same conclusions can be obtained as in the case where $\sin \frac{x + y}{2} = 0$, so the solution to the original system of equations is $\left\{\begin{array}{l}x = k_1 \pi \\ y = k_2 \pi \\ z = k_3 \pi .\end{array}\right.$