Solving arcsin(√2) with Hyperbolic Sin Function

[Curious3141]

actually wait one second... if I choose a=∏ , then my b=ln(-2±√5), which is a real solution for ln(-2+√5).

So do I have two different a and b combination solutions? i.e. if a=pi, then b=n(2+√5), and if a=0, b=(2+√5)?

You should get:

$a = 2k\pi, b = \ln(\sqrt{5} + 2)$

and

$a = (2k+1)\pi, b = \ln(\sqrt{5} - 2)$

for integral k.

as your general solution. The first solution set corresponds to even $n$, while the second, to odd $n$. As I said before, consider the cases where $n$ is even and odd separately when solving for $b$. It is not sufficient to consider only the special cases where $a = 0$ or $a = \pi$ unless the question explicitly gives you bounds for $a$.