Complex sine

Here's a really lazy method. In the example we already get the solution for $\cos{x} = 2$ , which is $x = 2k\pi \pm \ln{2 \pm \sqrt{3}}$ and $x = 2k\pi \mp \ln{2 \pm \sqrt{3}}$ . Fortunately, of the options provided, the part outside the logarithm has magnitude less than $2\pi$ , so we can drop all the $2k\pi$ to get $x = \pm \ln{2 \pm \sqrt{3}}$ or $x = \mp \ln{2 \pm \sqrt{3}}$ . Now, I used the complimentary angle identity: $\sin{x} = \cos{\frac{\pi}{2} - x}$ , letting $\sin{x} = 2$ : $\frac{\pi}{2} - x = \pm\mp \ln{2 \pm \sqrt{3}}$ , so we have $x = \frac{\pi}{2} \pm\mp \ln{2 \pm \sqrt{3}}$ . Check the options and only one match this form.

Are the options $\dfrac \pi 2 \pm i (\ln (2+\sqrt 3))$ and $\dfrac \pi 2 \pm i (\ln (2-\sqrt 3))$ identical?

The answer gets very complicated difficult for me to type in. A direct link to a website is given which provides the direct formula and its proof for these sin problems.

Can you solve sin x = 2?