If $x\sin A=y\sin(A+2π/3) =z\sin(A+4π/3),$ derive a relation among $x, y, z$ by eliminating $A$ 
If $x\sin A=y\sin(A+2π/3) =z\sin(A+4π/3),$ derive a relation among $x, y, z$ by eliminating $A$

This problem was bothering me for a while, and I finally could not solve it. I tried taking the whole equation as $k$ but the calculation was a mess. Would someone please help me to find a solution using a simpler approach? 
 A: Use $$\sin{A}+\sin(A+120^{\circ})+\sin(A+240^{\circ})=0.$$
Indeed, let $$x\sin{A}=y\sin(A+120^{\circ})=z\sin(A+240^{\circ})=k.$$
This, for $xyz\neq0$ and $k\neq0$ we obtain:
 $$\frac{k}{x}+\frac{k}{y}+\frac{k}{z}=0$$ or
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.$$
A: Consider (first & second terms) $$x\sin A=y\sin\left(A+\frac{2\pi}{3}\right)$$
$$x\sin A=y\left(\sin A\cos \frac{2\pi}{3}+\cos A\sin\frac{2\pi}{3}\right)$$
$$x=y\left(\frac{\sin A\cos \frac{2\pi}{3}}{\sin A}+\frac{\cos A\sin\frac{2\pi}{3}}{\sin A}\right)$$
$$x=y\left(-\frac12+\cot A \frac{\sqrt3}{2}\right)$$
$$\cot A=\frac{2}{\sqrt3}\left(\frac xy+\frac12\right)\tag 1$$
Similarly, consider (first & third terms) $$x\sin A=z\sin\left(A+\frac{4\pi}{3}\right)$$
$$x\sin A=z\left(\sin A\cos \frac{4\pi}{3}+\cos A\sin\frac{4\pi}{3}\right)$$
$$x=z\left(\frac{\sin A\cos \frac{4\pi}{3}}{\sin A}+\frac{\cos A\sin\frac{4\pi}{3}}{\sin A}\right)$$
$$x=z\left(-\frac12+\cot A \left(-\frac{\sqrt3}{2}\right)\right)$$
$$\cot A=-\frac{2}{\sqrt3}\left(\frac xz+\frac12\right)\tag 2$$
Equating (1) & (2), we get
$$\frac{2}{\sqrt3}\left(\frac xy+\frac12\right)=-\frac{2}{\sqrt3}\left(\frac xz+\frac12\right)$$
$$\frac xy+\frac xz+1=0$$
Hope above is the required relation among $x, y$ & $z$
