= Solution
Write $U=V_g$ for the coupling Fourier coefficient, with $|U|=|V_0|$ for the supplied cosine potential. Projecting the <Schrödinger equation> onto the two retained <plane waves> gives
$$
\begin{pmatrix}\epsilon_{q+g/2}&U\\U^*&\epsilon_{q-g/2}\end{pmatrix}
\begin{pmatrix}A\\B\end{pmatrix}
=E\begin{pmatrix}A\\B\end{pmatrix}.
$$
The diagonal potential average is zero. The <characteristic equation> is
$$
(E-\epsilon_{q+g/2})(E-\epsilon_{q-g/2})-|U|^2=0,
$$
so the <nearly-free electron dispersion near a one-dimensional band gap> is
$$
\boxed{E_\pm(q)=\frac{\epsilon_{q+g/2}+\epsilon_{q-g/2}}2
\pm\frac12\sqrt{(\epsilon_{q+g/2}-\epsilon_{q-g/2})^2+4V_0^2}.}
$$
Only the coupling magnitude enters the energies; its sign changes the standing-wave eigenvectors. The two-wave truncation is controlled near this crossing when $|V_0|\ll\epsilon_{g/2}$, since the other <plane waves> are separated by much larger kinetic energies.
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