<s>[INST] How do we solve $(w^2-w^{-2})(2w^2+2w^{-2} -1)-\sqrt{3}i(w^2+w^{-2})=0$? How do we solve the following equation?
$$(w^2-w^{-2})(2w^2+2w^{-2} -1)-\sqrt{3}i(w^2+w^{-2})=0$$
where $w = e^{ix}$.
UPDATE: 
I converted the trigonometric equation $2\sin(4x) -\sin(2x) - \sqrt{3}\cos(2x) = 0$ to euler. 
 [/INST] We have
$$
\sin(4x)=\sin\left(2x+\frac{\pi}{3}\right)
$$
So
$$
4x = n\pi + (-1)^n \left(2x+\frac{\pi}{3}\right)
$$
and hence
$$
x = \frac{n\pi + (-1)^n \frac{\pi}{3}}{2(2-(-1)^n)}
$$

If you don't want to use Euler, note that the given equation is
$$
2 w^8 - (1+i \sqrt{3}) w^6 - (i \sqrt{3}-1) w^2  - 2 = 0
$$
The LHS has an obvious factor $2w^2-(1+i\sqrt{3})$, and so
$$
[2w^2-(1+i\sqrt{3})] \left( w^6 - \frac{1+i \sqrt{3}}2\right)=0
$$
so you know the eight roots of $w$.
</s>