Is this trig even real?

For this, we need to use the complex definition of the sine function (derived from Euler's Formula ):

$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$

We can set this to be $2$ and then we can neaten it up:

$\frac{e^{iz}-e^{-iz}}{2i}=2 \Longrightarrow e^{iz}-e^{-iz}=4i$

We can now multiply both sides by $e^{iz}$ and we can form a quadratic equation:

$e^{iz}-e^{-iz}=4i \Longrightarrow \left(e^{iz}\right)^2-1=4ie^{iz} \Longrightarrow \left(e^{iz}\right)^2-4ie^{iz}-1=0$

Now, using the Quadratic Formula , we can determine that, with $a=1, b=-4i, c=-1$ :

$e^{iz}=\frac{-(-4i)\pm\sqrt{(-4i)^2-4(1)(-1)}}{2(1)} \Longrightarrow e^{iz}=\frac{4i\pm\sqrt{-12}}{2} \Longrightarrow e^{iz}=i(2\pm\sqrt{3})$

Now, we can take the natural log to get:

$iz=\ln(i(2\pm\sqrt{3}) \Longrightarrow iz=\ln(i)+\ln(2\pm\sqrt{3})$

Since we know that $\ln(i)=e^{i\frac{\pi}{2}}$ and we are only taking the principle value we can say that

$iz=i\frac{\pi}{2}+\ln(2\pm\sqrt{3})$

Now we can divide by $i$ to find that

$\boxed{z=\frac{\pi}{2}-\ln(2\pm\sqrt{3})i}$