If $\mathrm{xsin}\mathrm{a}+\mathrm{ysin}2\mathrm{a}+\mathrm{zsin}3\mathrm{a}=\sin 4\mathrm{a}; \mathrm{xsin}\mathrm{b}+\mathrm{ysin}2\mathrm{b}+\mathrm{zsin}3\mathrm{b}=\sin 4\mathrm{b}, \mathrm{xsin}\mathrm{c}+\mathrm{ysin}2\mathrm{c}+\mathrm{zsin}3\mathrm{c}=\sin 4\mathrm{c}$. Then the roots of the equation ${\mathrm{t}}^3-\left(\frac{\mathrm{z}}{2}\right){\mathrm{t}}^2-\left(\frac{\mathrm{y}+\mathrm{z}}{4}\right)\mathrm{t}+\left(\frac{\mathrm{z}-\mathrm{x}}{8}\right)=0,\mathrm{a},\mathrm{b},\mathrm{c}\ne \mathrm{n\pi }$, are

Equation first can be written as

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

Similarly, from second and third equation we can verify cos b and cos c are the roots of the given equation