Solutions of $x^2=1$ Show that in a cyclic group, the equation $x^2=1$ has no more than two solutions.

Since $G$ is a cyclic group, an element $x\in G$ can be written as $\langle x \rangle=\{x^n\mid n\in\mathbb{Z}\}=G$. Since $x^2=1$, the order of $x$ is $2$. Since there are only two positive divisors of $2$ (namely $1$ and $2$), there can be no more than two solutions to $x^2=1$ in a cyclic group.
I'm not sure if my reasoning is correct for this problem. Can someone let me know if I'm going in the right direction?

Hint: Let $g$ be the generator for the group, and write out all the elements of the group as $1, g, g^2, \cdots, g^{n-1}$.  Then $x$ has to be one of these, so let $x = g^k$ for $0 \le k \le n-1$.
If the cyclic group is infinite, then write it out as $\ldots, g^{-2}, g^{-1}, 1, g, g^2, \ldots$ and do the same thing.
