Am I correct in saying that sinh(x) = 0 only when x = 0 and cosh(x) never equals 0?
Yes. To prove it, write $\displaystyle \displaystyle \sinh{x} = \frac{e^x - e^{-x}}{2}$, set it equal to 0 and solve for $\displaystyle \displaystyle x$. Also, set $\displaystyle \displaystyle \cosh{x} = \frac{e^x + e^{-x}}{2}$, and show that this is always positive.