#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 $n \in \mathbb{Z}$:

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