<s>[INST] Solve trigonometric function $x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0$ I need to solve a trigonometric function similar to the following one for $\alpha$.
$$
x_1 \sin(2\alpha)+x_2 \cos(2\alpha) - x_3 \sin(\alpha) - x_4 \cos(\alpha) = 0 
$$
I found a solution to a very similar problem
$$\frac{1}{2} \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0. $$
here, but the $\cos(2\alpha)$ term causes me some trouble, to adapt this solution. 
TIA for your time!
 [/INST] Using the complex representation, we have $z:=e^{i\alpha}$, $\cos\alpha=\frac{z+z^{-1}}2$, $\sin\alpha=\frac{z-z^{-1}}{2i}$, and $\cos2\alpha=\frac{z^2+z^{-2}}2$, $\sin2\alpha=\frac{z^2-z^{-2}}{2i}$.
$$x_1\frac{z^2-z^{-2}}{2i}+x_2\frac{z^2+z^{-2}}2-x_3\frac{z-z^{-1}}{2i}-x_4\frac{z+z^{-1}}2=0,$$
or
$$x_1\frac{z^4-1}{2i}+x_2\frac{z^4+1}2-x_3\frac{z^3-z}{2i}-x_4\frac{z^3+z}2=0.$$
This is a general quartic equation in $z$ and unless the coefficients have special values, there is no easy solution.
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