sin x = 2

Math Trig

$$\begin{gathered} \sin x=2 \\ \frac{e^{ix}-e^{-ix}}{2i}=2 \\ {e}^{ix}-e^{-ix}=4i \\ {e}^{ix}-(e^{ix}{)}^{-1}=4i \end{gathered}$$

Let $u=e^{ix}$:

$$\begin{gathered} u-u^{-1}=4i \\ {u}^2-1=4iu \\ {u}^2-4iu-1=0 \\ {u}^2-4iu-4=-3 \\ (u-2i{)}^2=-3 \\ u-2i=\pm \sqrt{-3} \\ u=2i\pm \sqrt{-3} \\ u=i(2\pm \sqrt{3}) \end{gathered}$$

Substitute back into $u$, for $n\in \mathbb{Z}$:

$$\begin{gathered} e^{ix}=i(2\pm \sqrt{3}) \\ ix=\ln (i(2\pm \sqrt{3})) \\ ix=\ln i+2\pi n+\ln (2\pm \sqrt{3}) \\ ix=\frac{i\pi }{2}+2\pi n+\ln (2\pm \sqrt{3}) \\ x=\frac{\pi }{2}-i\ln (2\pm \sqrt{3})+2\pi n \end{gathered}$$