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 half-integer $\sin$'s
\begin{equation}\label{error}\lambda_n=-\Bigl(n+\frac{1}{2}\Bigr)^2, X_n(x)=\sin \Bigl(n+\frac{1}{2}\Bigr)x,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 $A=0$ and
$$\gamma B\cosh \gamma\pi+\alpha B\sinh \gamma\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=\gamma_n^2$ where $\gamma_n$ is a nonzero root of
$$\gamma=-\alpha\tanh \gamma\pi.$$
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 $A=0$ and
$$\omega B\cos \omega\pi+\alpha B\sin \omega\pi=0.$$
Hence the various eigenvalues are given by $\lambda_n=-\omega_n^2$ where $\omega_n$ is a nonzero root of
$$\omega=-\alpha\tan \omega\pi.$$
If $\lambda=0$ then we have only trivial solution.
