How to find the $\arcsin 2$?

This is an answer that gives imaginary solutions. Note that the OP never specified that this couldn't be used.

You are trying to find:

$$\sin x=2$$

This obviously has no real solutions, so we will look for imaginary solutions. Rewrite $x$ as $a+bi$ where $a,b\in\Bbb{R}$. That is crucial.

$$\sin (a+bi)=2$$

Using the arguement sum property (not sure of the name) of $\sin$, you can isolate the $a$ and $bi$ into bite-sized parts. Also, multiply $-i$ and $i$

$$\sin a \cos bi+\cos a \sin bi=2$$

Now, we know some properties of hyperbolic trig allowing us to alter some of the stuff along the following rules. Those rules would be here. Now we are getting somewhere.

$$\sin a \cosh b+(\cos a) *(i\sinh b)=2$$ which we can rewrite as

$$\sin a \cosh b+i\cos a\sinh b=2$$

Because $a,b$ have to be real, and that $\sinh b\not=0$, you are forced to assume that $a=\pi/2+2\pi z$. The reason? Well, the term $i\cos a \sinh b$ is imaginary and the term (2) you are trying to get is strictly real. It is imaginary because both $\cos$ and $\sinh $ are functions of $\Bbb{R}\to\Bbb{R}$

So, we now know that $a=\pi/2+2\pi z$ from logical sense. (where $z\in\Bbb{Z}$)

$$\sin(\pi/2+2\pi z)\cosh b=2\implies \cosh b=2$$

Now since $\cosh$ is an even function:

$$b=\pm \text{arcosh}(2)$$

Plugging that back into the almost original equation, the one with $\sin(a+bi)$, we get:

$$\sin(\frac{\pi}{2}+2\pi z\pm i\text{arccosh}(2) )=2$$

Viola!

You don't. Arcsine is the inverse function of sine. The domain of arcsine is the range of sine which is $[-1,1]$ and $2$ is not in there.

The arcsin(2) is not a real number. Recall that $$x = \arcsin(2)$$ is equivalent to the equation $$\sin(x) = 2.$$ Since the range of $\sin(x)$ as a real valued function is $[-1,1]$, the original equation has no real solution. I doubt a calculus exam wants a complex valued solution.