What's the value of x when sin⁡x=2 and x is a complex number?

I like these. I think I did ${\displaystyle \sin x=8}$ in a previous answer.
${\displaystyle e^{ix}=\cos x+i\sin x}$
${\displaystyle e^{ix}=\cos x-i\sin x}$
${\displaystyle e^{ix}-e^{-ix}=2i\sin x}$
${\displaystyle \sin x=\frac{e^{ix}-e^{-ix}}{2i}}$
Let ${\displaystyle y=e^{ix}}$. Then ${\displaystyle e^{-ix}=\frac{1}{y}}$.
${\displaystyle 2\left(2i\right)=y-\frac{1}{y}}$
${\displaystyle y^2-4iy-1=0}$
${\displaystyle y=2i\pm \sqrt{-3}=i(2\pm \sqrt{3})}$
Convert to polar and get everything in the exponent. We note ${\displaystyle e^{\frac{i\pi }{2}}=i\text{and}{e}^{2\pi ki}=1}$ for integer k.
${\displaystyle e^{ix}=y=e^{\frac{i\pi }{2}}{e}^{\ln (2\pm \sqrt{3})}{e}^{2\pi ki}}$
${\displaystyle ix=i\frac{\pi }{2}+\ln (2\pm \sqrt{3})+2\pi ki}$
${\displaystyle x=\frac{\pi }{2}+2\pi k-i\ln (2\pm \sqrt{3})}$ for integer k
Check: Let’s check one, plus root, ${\displaystyle k=0}$.
${\displaystyle x=\frac{\pi }{2}-i\ln (2+\sqrt{3})}$
${\displaystyle e^{ix}=e^{\frac{i\pi }{2}}(2+\sqrt{3})=i(2-\sqrt{3})}$
${\displaystyle e^{-ix}=\frac{1}{i(2-\sqrt{3})}=-i(2+\sqrt{3})=i(-2-\sqrt{3})}$
${\displaystyle e^{ix}-e^{-ix}=4i}$
${\displaystyle \sin x=\frac{e^{ix}-e^{-ix}}{2i}=2}$

${\displaystyle \sin x=2}$
We use the fact that
${\displaystyle \sin x=\frac{e^{ix}-e^{-ix}}{2i}}$
So our equation is
${\displaystyle \frac{e^{ix}-e^{-ix}}{2i}=2}$
Now, set ${\displaystyle z=e^{ix}}$. So we have
${\displaystyle \frac{z-\frac{1}{z}}{2i}=2}$
which can be rearranged as
${\displaystyle z^2-4iz-1=0}$
Now, solve this
${\displaystyle {(z-2i)}^2+4-1=0}$
in other words
${\displaystyle {(z-2i)}^2-{\left(i\sqrt{3}\right)}^2=0}$
which is also
${\displaystyle (z-(2+\sqrt{3})i)(z-(2-\sqrt{3})i)=0}$
So we have two solutions:
${\displaystyle z_1=(2+\sqrt{3})i\text{and}{z}_2=(2-\sqrt{3})i}$
Now, let’s get back to our initial variable x.
${\displaystyle e^{ix}=(2+\sqrt{3})i}$
implies that
${\displaystyle x=-i\ln i-i\ln (2+\sqrt{3})+2k\pi ,(k\in Z)}$
${\displaystyle =2k\pi +\frac{\pi }{2}-i\ln (2+\sqrt{3})}$
Likewise, we find that ${\displaystyle x^{\prime }=2k\pi +\frac{\pi }{2}-i\ln (2-\sqrt{3})}$ are the other class of solutions.
Hence, the solutions are ${\displaystyle (2k+\frac{1}{2})\pi -i\ln (2\pm \sqrt{3})}$ where ${\displaystyle k\in Z}$.