<s>[INST] Solving $\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$ where $x$, $y$, $z$ are angles of a triangle Can any one give me a hint to find value of $x$
where: 

$$\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$$
  and $x$, $y$, $z$ are angles of a triangle.

I tried to use sine law but got nothing.
 [/INST] Unless $x, y, z$ represent the angles of a triangle, there'll be infinitely many solutions.
If three sides of a triangle are given, it's better to use cosine law:
\begin{align*}
  \cos x &= \frac{3^{2}+2^{2}-4^{2}}{2(3)(2)} \\
  \cos y &= \frac{4^{2}+2^{2}-3^{2}}{2(4)(2)} \\
  \cos z &= \frac{3^{2}+4^{2}-2^{2}}{2(3)(4)} \\
\end{align*}
</s>