The associated eigenproblem is
\begin{equation}\left\{\begin{split}&X''=\lambda X,\\&X'|_{x=0}=(X'-\alpha X)|_{x=0}=0.\end{split}\right.\label{1-4}\end{equation}
If $\alpha=0$ the we know the solution are integer $\cos$'s
\begin{equation}\label{error}\lambda_n=-n^2, X_n(x)=\cos nx, n=0,1,....\end{equation}
If $\alpha\neq 0,\lambda>0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cosh \gamma x + B\sinh \gamma x, \gamma>0.$$
Plugging in boundary condition we find $B=0$ and
$$\gamma A\sinh \gamma\pi+\alpha A\cosh \gamma\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=\gamma_n^2$ where $\gamma_n$ is a nonzero root of
$$\gamma\tanh \gamma\pi+\alpha=0.$$
If $\alpha\neq 0,\lambda<0$ then the general solution for the DE in \eqref{1-4} is
$$X(x)=A\cos \omega x + B\sin \omega x, \omega>0.$$
Plugging in boundary condition we find $B=0$ and
$$-\omega A\sin\omega\pi+\alpha A\cos \omega\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=-\omega_n^2$ where $\omega_n$ is a nonzero root of
$$\omega\tan \omega\pi-\alpha=0.$$
If $\lambda=0$ then we have only trivial solution.
