What are the solutions for $\sinh(y)$ $=$ $(-1)^n$ $\sinh(1)$? What I thought first is, assuming two cases
Case I: $\sinh(y)$ $=$ $\sinh(1)$
$=>$ $y=1$
Case II: $\sinh(y) = -\sinh(1) = \sinh(-1) => y = -1$
Is this correct way of solving? Or should I convert them into exponential function and solve for $y$?
 A: Yes, what you have written is correct. You can also solve the problem using exponentials, but what you have done is probably simpler. Here  is my attempt to give some further justification for the steps that you have used.
The function $n\mapsto(-1)^n$ is only defined when $n$ is a rational number with an odd denominator (assuming that you are solving this problem over the real numbers). If the numerator of $n$ is even, $(-1)^n=1$, and if the numerator is odd, $(-1)^n=-1$. So the function $n\mapsto(-1)^n$ only takes on the values of $-1$ and $1$, meaning that it is valid to split the problem into two different cases:

*

*Case 1: $(-1)^n=1$.

*Case 2: $(-1)^n=-1$.

In case 1, the equation is $\sinh y=\sinh 1$. Noting that $\sinh$ is strictly increasing on $\Bbb{R}$, and so is one-to-one, we get that $\sinh y=\sinh 1  \implies y=1$.
In case 2, the equation is $\sinh y=-\sinh1$. Since $\sinh$ is odd, $\sinh(-1)=-\sinh1$, and so what you have written is correct.
Hence, if $\sinh y=(-1)^n\sinh1$, then $n$ has an even numerator and $y=1$, or $n$ has an odd numerator and $y=-1$.
