Solve for $x$ using Euler Identity.
$$\tan(2x) = -1$$
We can first note that
$$\tan(2x) = \dfrac{\sin(2x)}{\cos(2x)} = \dfrac{e^{2ix}-e^{-2ix}}{i(e^{2ix}+e^{-2ix})} = -1$$
$$e^{2ix}- e^{-2ix}+i(e^{2ix}+e^{-2ix}) = 0$$
Let $w = e^{2ix}$
$$w-w^{-1} +i(w+w^{-1}) = 0$$
$$w^2 -1+i(w^2+1) = 0$$
This is where I'm stuck.
Note that$$w^2-1+i(w^2+1)=0\iff w^2=-i.$$So, take the square roots of $-i$, which are $\pm\left(-\frac1{\sqrt2}+\frac i{\sqrt2}\right)$.