sin x = 2

Math Trig

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

Let $u=e^{ix}$:

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

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

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