Solving Wave Equation on $0 < x < L$ with initial condition containing both sine and cosine terms I am trying to solve the wave equation as follows:
$$u_{tt} = u_{xx} \,\,\,\,\,\,\,\, 0 < x < L$$
$$u(0,t) = u(L,t)$$
$$u_x(0,t) = u_x(L,t)$$
$$u(x,0) = sin\bigg(\frac{2\pi x}{L}\bigg) + cos\bigg(\frac{4\pi x}{L}\bigg)$$
The issue I'm running into is somehow I end up with $-1 = 1$ and I'm not sure what I'm doing wrong.
I use separation of variables and get 
$$X(x) = a\,\text{cos}(\lambda x) + b\,\text{sin}(\lambda x).$$ The boundary conditions for $u$ imply $X(0) = X(L)$ so I set these equal to each other and get that the coefficient $a$ is $$ a = \frac{b\,\text{sin}(\lambda L)}{1 - cos(\lambda L)}.$$ This then yields
$$X(x) = \frac{b\,\text{sin}(\lambda L)}{1 - cos(\lambda L)}\,\text{cos}(\lambda x) + b\,\text{sin}(\lambda x)$$
I'm going to omit all the calculations because it's kind of messy, but my process is as follows: 
The boundary conditions also imply $X'(0) =  X'(L)$ so I plug each one into the derivative of X, $X'(x)$. Then I set these equal to each other and solve. I end up with $-1 = 1$. This is obviously incorrect. Can someone point me in the right direction? Also obviously didn't even start to solve for $T(t)$ yet. 
 A: You were going in the right direction. The mistake is probably in your algebra somewhere. Since you didn't show your work (please edit to add that in if you can), I can't say more. 
The corresponding derivatives are
\begin{align}
X'(0) &= \lambda b \\
X'(L) &= -\frac{\lambda b\sin(\lambda L)}{1-\cos(\lambda L)}\sin(\lambda L) + \lambda b \cos(\lambda L)
\end{align}
Assuming $b \ne 0$, one solution is obviously $\lambda = 0$. If $\lambda \ne 0$, the equation reduces to
$$ 1 = -\frac{\sin^2(\lambda L)}{1-\cos(\lambda L)} + \cos(\lambda L) $$
Going through the steps
$$ 1-\cos(\lambda L) = -\sin^2(\lambda L) + \cos(\lambda L)\big(1-\cos(\lambda L)\big) $$
$$ \implies 1 - \cos(\lambda L) = \cos(\lambda L) - 1 $$
$$ \implies \cos(\lambda L) = 1 $$
$$ \implies \lambda L = 2n\pi, \quad n = 0,1,2,\dots $$
Using the property of linear equations, we obtain a series solution:
$$ u(x,t) = a_0(c_0 + d_0 t) \\ + \sum_{n=1}^\infty \left[a_n \cos\left(\frac{2n\pi}{L} x\right) + b_n \sin\left(\frac{2n\pi}{L} x\right)\right]\left[c_n \cos\left(\frac{2n\pi}{L} t\right) + d_n \sin\left(\frac{2n\pi}{L} t\right)\right] $$
We can sort of simplify this by break up the terms and renaming the constants
$$ u(x,t) = A_0 + B_0 t + \sum_{n=1}^\infty \left[A_n \cos\left(\frac{2n\pi}{L} x\right) + B_n \sin\left(\frac{2n\pi}{L} x\right)\right]\cos\left(\frac{2n\pi}{L} t\right)\\ + \left[C_n \cos\left(\frac{2n\pi}{L} x\right) + D_n \sin\left(\frac{2n\pi}{L} x\right)\right]\sin\left(\frac{2n\pi}{L} t\right)$$
Then, you'll need to solve 2 different Fourier series
\begin{align} 
u(x,0) &= f(x) = A_0 + \sum_{n=1}^\infty \left[A_n \cos\left(\frac{2n\pi}{L} x\right) + B_n \sin\left(\frac{2n\pi}{L} x\right)\right] \\ 
u_t(x,0) &= g(x) = B_0 + \sum_{n=1}^\infty \frac{2n\pi}{L}\left[C_n \cos\left(\frac{2n\pi}{L} x\right) + D_n \sin\left(\frac{2n\pi}{L} x\right)\right]
\end{align}
