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:

$$

and

$$

for integral k.

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