Use Fourier's technique to solve the general wave equation
$$
\begin{array}{ll}
u_{tt} = c^2 u_{xx} & u(0,t)=u(L,t)=0 \\
u(x,0) = f(x) & u_t(x,0) = g(x).
\end{array}
$$
You answer should be expressed as an infinite sum whose coefficients are expressed as integrals involving the functions $f$ and $g$.
We can reasonably assume $\text{u(x,t) = }
c^2 X(x)T(t)$ and use separation of variables using $-\lambda ^2$ as a constant. We then simply have two ODEs with solutions:
$$\text{X(x) = }
A \sin \left(\frac{\lambda x}{c}\right)+B \cos \left(\frac{\lambda x}{c}\right)
$$$$
\text{
T(t) = }C \sin (\text{$\lambda $t})+D \cos (\text{$\lambda $t})$$
We can now apply the boundary conditions: $u(0,t)=u(L,t)=0$. First the left boundary: $$u(0,t)=0$$$$x(0)=B cos(\frac{\lambda x}{c})=0 $$
Ignoring the trivial case, necessarily $B=0$. Next the right boundary:
$$u(0,L)=0$$$$X(L) = A \sin \left(\frac{\lambda x}{c}\right)+0=0$$ This gives us the following eigenvalues for lambda:
$$\lambda =\frac{\text{n$\pi $c}}{L}$$
And:
$$X(x)=\frac{A \sin (\lambda x)}{c}$$
We can express $u(x,t)$ with coefficients $b_n$ and shown below with its derivative (which we will use later)
$$\text{u(x,t)=}
\sum _{n=1}^{\infty } b_n T(t) X(x)$$$$\text{$\Rightarrow $}
\sum _{n=1}^{\infty } A b_n \sin \left(\frac{\lambda x}{c}\right) (C \sin (\text{$\lambda $t})+D \cos (\text{$\lambda $t}))$$$$u_t
\text{(x,t)=}
\sum _{n=1}^{\infty } A \lambda b_n \sin \left(\frac{\lambda x}{c}\right) (C \cos (\text{$\lambda $t})-D \sin (\text{$\lambda $t}))$$
Applying the boundary conditions for $u(x,t)$ we get:
$u(x,0)=f(x)=\sum _{n=1}^{\infty } {A D b_n \sin (\frac\lambda{c} x)}$
$u_t(x,0)=g(x)=\sum _{n=1}^{\infty } \text{AC} \lambda b_n \sin \left(\frac{\lambda x}{c}\right)$
Let $\alpha_n=b_nAD$ and $\beta_n=b_n\lambda AC$ just to tidy up. Since our bound is from $0$ to $L$ we have a factor of two outside the following integrals for $\alpha_n$ and $\beta_n$ with our expression for $u(x,t)$
$$\text{u(x,t)=}
\sum _{n=1}^{\infty } A b_n \sin \left(\frac{\lambda x}{c}\right) (C \sin (\text{$\lambda $t})+D \cos (\text{$\lambda $t}))$$$$\alpha _n\text{= }\frac{2}{L}\int_0^L f(x) \sin \left(\frac{\lambda x}{c}\right) \, dx$$$$
\beta _n\text{= }\frac{2}{L}\int_0^L g(x) \sin \left(\frac{\lambda x}{c}\right) \, dx$$
where:
$\alpha_n=b_nAD$
$\beta_n=b_n\lambda AC$
*I apologize for any LaTeX related errors, they will be corrected as found.
So I liked the way you did everything up until the final statement of $u(x,t)$. Instead of leaving it in terms of $A$, $B$, and $C$ which we don't have specific equations for, why not leave it in the form of $\alpha$, and $\beta$ which we do have?
$$u(x,t)=\sum_{n=1}^{\infty}\sin\left(\frac{\lambda x}{c}\right)\left(\alpha_n\cos(\lambda t)+\left(\frac{\beta_n}{\lambda}\right)\sin(\lambda t)\right)$$