You are trying to find the general solution, which is impossible. However what is required? To find solutions which are travelling waves that means

\begin{equation}

u= \phi (x-vt)

\label{A}

\end{equation}

with unknown function $\phi$ and constant $v$. Then $u_{xx}=phi ''$, $u_{tt}=v^2 \phi''$ and we get equation

\begin{equation}

(v^2-c^2) \phi '' + \sin (\phi) =0.

\label{B}

\end{equation}

This is ODE and we are looking for its bounded solutions. When $v =\pm c$ we get constant solutions $\phi(\xi)=\pi n$ (which are definitely of no interest).

Let $v^2>c^2$. Then we get equation describing mathematical pendulum $\phi '' + k\sin (\phi)=0$

https://en.wikipedia.org/wiki/Pendulum_(mathematics) which you definitely considered in the ODE class at Chapter 9 of Boyce and Di Prima, $k=g/L$ where $L$ is a distance from the anchor to the center of mass ($\phi$ is a deviation from the stable equilibrium $2\pi n$). While explicit solution is impossible we know that such equation has periodic (with respect to $\xi$) solutions which we are interested in (there are also non-periodic solutions corresponding pendulum having so large speed that it passes through the top point but we are not interested in those).

Let $v^2<c^2$. Then $k<0$ but replacing $\phi$ by $\phi+\pi$ we get the same equation albeit with $k$ replaced by $-k$.

So, conclusion which we are looking:

For each $v\ne \pm c$ equation $u_{tt}-c^2 u_{xx} + \sin (u)=0$ has solutions of the form (\ref{A}) with periodic function $\phi$ (satisfying (\ref{B}).

Question: which periods can have $\phi$? Since $\phi''+ k\sin(\phi)=0$ has solution for any period $T\in (2\pi/\sqrt{k},\infty)$, we have all periods in the interval ${(2\pi |v^2-c^2|^{-1/2},\infty)}$.