$i(w^2 +w^{-2})-(w-w^{-1}) = i$

Question

How do we solve the following equation?

$$i(w^2 +w^{-2})-(w-w^{-1}) = i$$

where $w = e^{ix}$ and $i$ is a complex number.

Answer 1

You have $\cos 2x-\sin x=\frac12$, and recall double angle formula for cosine.

Alternatively, note that $w^2+w^{-2}=(w-w^{-1})^2+2$, and solve the quadratic in $w-w^{-1}$, hence $w$.