If x sin a +y sin 2a + z sin3a = sin4a,  x sin b +y sin 2b + z sin3b= sin4b,  x sin c+ y sin 2c +z sin 3c= sin4c, then the roots of the equation

${\mathrm{t}}^3-\left(\frac{\mathrm{z}}{2}\right){\mathrm{t}}^2-\left(\frac{\mathrm{y}+2}{4}\right)\mathrm{t}+\left(\frac{\mathrm{z}-\mathrm{x}}{8}\right)=0,\mathrm{a},\mathrm{b},\mathrm{c}\ne \mathrm{n\pi }$ are

$\begin{array}{l}\mathrm{xsin}\mathrm{a}+\mathrm{ysin}2\mathrm{a}+\mathrm{zsin}3\mathrm{a}=\sin 4\mathrm{a}\\ \Rightarrow \mathrm{xsin}\mathrm{a}+\mathrm{y}\times 2\sin \mathrm{acos}\mathrm{a}+\mathrm{z}\times \sin \mathrm{a}\left(3-4{\sin }^2\mathrm{a}\right)\\ =2\times 2\sin \mathrm{acos}\mathrm{acos}2\mathrm{a}\\ \Rightarrow \mathrm{x}+2\mathrm{ycos}\mathrm{a}+\mathrm{z}\left(3+4{\cos }^2\mathrm{a}-4\right)\\ =4\cos \mathrm{a}\left(2{\cos }^2\mathrm{a}-1\right)[\text{ as }\sin \mathrm{a}\ne 0]\end{array}$

$\begin{array}{l}\Rightarrow 8{\cos }^3\mathrm{a}-4{\mathrm{zcos}}^2\mathrm{a}-(2\mathrm{y}+4)\cos \mathrm{a}+(\mathrm{z}-\mathrm{x})=0\\ \Rightarrow {\cos }^3\mathrm{a}-\left(\frac{\mathrm{z}}{2}\right){\cos }^2\mathrm{a}-\left(\frac{\mathrm{y}+2}{4}\right)\cos \mathrm{a}+\left(\frac{\mathrm{z}-\mathrm{x}}{8}\right)=0\end{array}$

Which shows that cos a is root of the equation 

${\mathrm{t}}^3-\left(\frac{\mathrm{z}}{2}\right){\mathrm{t}}^2-\left(\frac{\mathrm{y}+2}{4}\right)\mathrm{t}+\left(\frac{\mathrm{z}-\mathrm{x}}{8}\right)=0$

Similarly, from second and third equations, we can slow that cos b and cos c are the roots of the given equation.