Vibrating string - separation of variables $u_{tt}=c^2u_{xx}$ where $u(x,0)=x+\sin(x)$, $u_t(x,0)=0$, $u(0,t)=u_x(\pi,t)=0$.
Assume a solution $u(x,t)=X(x)T(t)\not\equiv 0$.  This yielded $\lambda_n=\frac{1}{2}+2n$.  For $X_n(x)$ I have $$X_n(x)=B_n\left(\frac{1}{2}+2n\right)\cos\left(\left(\frac{1}{2}+2n\right)x\right)$$
Now I'm attempting to solve the equation $$T''-\left(\frac{1}{2}+2n\right)c^2=0.$$
For $T$ we have:  $$T(t)=C\cos\left(\left(\frac{1}{2}+2n\right)ct\right)+D\sin\left(\left(\frac{1}{2}+2n\right)c t\right).$$
Applying our conditions gives us: $x+\sin(x)=C$ and $ 0=D\left(\frac{1}{2}+2n\right)c.$
I'm again stuck trying to put together my $X_nT_n$.  But I'm also wary of my solution thus far.
 A: We know that: $$u=F(x)G(t)$$
and that we have $$F(x)\ddot{G}(t)=c^2F''(x)G(t)$$
So by separation of variables we obtain:
$$c^2\frac{F''(x)}{F(x)}=\frac{\ddot{G}(t)}{G(t)}=0$$
Since both sides are entirely dependent on one variable each, and they equal zero, we know they are constant. E.g.
$$c^2\frac{F''(x)}{F(x)}=\frac{\ddot{G}(t)}{G(t)}=k$$
Now we can treat them as second order differential equations and apply the boundary conditions.
Now this is definitely a problem of Fourier Analysis, and we can see that:
Boundary conditions: $u(0,t)=0$ but we don't have $u(\pi,t)=0$ but rather $u_x(\pi,t)=0$
Answer still in production...
A: $u(x,t)=X(x)T(t)$ leads to the eigenvalue problem $$X''+\lambda X=0,\quad X(0)=0,\ X'(\pi)=0 \tag{1}$$ and the time problem $$T''+\lambda c^2 T=0.\tag{2}$$
The eigenvalues are $$\lambda_n=\left(\frac{(2n-1)}{2}\right)^2=\left(n-\frac{1}{2}\right)^2, \quad n=1,2,\dots$$ with associated eigenfunctions $$X_n(x)=\sin\left(\sqrt{\lambda_n}\,x\right), \quad n=1,2,\dots$$
Now that the eigenvalues are known, we can solve $(2)$:
$$
T_n(t)=a_n\cos(\sqrt{\lambda_n}\,c\,t)+b_n\sin(\sqrt{\lambda_n}\,c\,t), \quad n=1,2,\dots
$$
Thus, the solution to the PDE and BCs is
\begin{align}
u(x,t)&=\sum_{n=1}^\infty X_n(x)T_n(t)\\
&=\sum_{n=1}^\infty \sin\left(\sqrt{\lambda_n}\,x\right)\left[a_n\cos(\sqrt{\lambda_n}\,c\,t)+b_n\sin(\sqrt{\lambda_n}\,c\,t)\right].\tag{3}
\end{align}
Now is when you apply the initial conditions (you did so too soon before) in order to determine the coefficients $a_n$ and $b_n$:
$$
u(x,0)=\sum_{n=1}^\infty a_n\sin\left(\sqrt{\lambda_n}\,x\right)=x+\sin(x), \quad 0<x<\pi,
$$
This is just a Fourier sine series for the function $x+\sin x$ on $0<x<\pi$, so
$$a_n={2\over \pi}\int_0^\pi(x+\sin x)\sin(\sqrt{\lambda_n}\,x)\,dx, \quad n=1,2,\dots
$$
To apply other initial condition, differentiate $(3)$ with respect to $t$, then evaluate it at $t=0$, and set the result equal to $0$ (since $u_t(x,0)=0$ was given):
$$
u_t(x,0)=\sum_{n=1}^\infty \sqrt{\lambda_n}\,c\,b_n\sin\left(\sqrt{\lambda_n}\,x\right)=0, \quad 0<x<\pi,
$$
Again, this is just a Fourier sine series for the function $0$ on $0<x<\pi$, so
\begin{align}
\sqrt{\lambda_n}\,c\,b_n&={2\over \pi}\int_0^\pi 0\cdot\sin(\sqrt{\lambda_n}\,x)\,dx, \quad n=1,2,\dots\\
&=0
\end{align}
so $b_n=0$, $n=1,2,\dots$
