= Solution
For potential period $a$, a <Bloch state> has the form $\psi_k(x)=e^{ikx}u_k(x)$, with $u_k(x+a)=u_k(x)$, or equivalently $\psi_k(x+a)=e^{ika}\psi_k(x)$. Quasimomentum is defined modulo the reciprocal-lattice vector $2\pi/a$. A <Brillouin zone> is one fundamental interval for that identification, usually $[-\pi/a,\pi/a)$. Here $a=2\pi$, so it is $[-1/2,1/2)$.
For small $V_0>0$, the first free-particle degeneracy occurs at the zone edge $k=1/2$, between <plane waves> $e^{ix/2}$ and $e^{-ix/2}$ of energy $1/4$. In this two-dimensional subspace, the diagonal <matrix> elements of the potential vanish. Its <Fourier coefficient> at wave number one is $-V_0/2$; the $\cos2x$ term cannot connect these two waves at first order. Thus <degenerate perturbation theory> gives the <matrix>
$$
\begin{pmatrix}1/4&-V_0/2\\-V_0/2&1/4\end{pmatrix},
$$
with energies $1/4\pm V_0/2+O(V_0^2)$. Near the edge the two uncoupled energies have opposite slopes, and diagonalization gives the avoided crossing, so the lowest <band gap> is
$$
\boxed{\Delta E=V_0+O(V_0^2),\qquad V_0\downarrow0.}
$$
For large $V_0$, expand about each deepest minimum $x=2\pi j$:
$$
-V_0(\cos x+\cos2x)=-2V_0+\frac52V_0x^2+O(V_0x^4).
$$
The local <quantum harmonic oscillator> $-d^2/dx^2+(5V_0/2)x^2$ has levels $(2n+1)\sqrt{5V_0/2}$. Its localization width is $O(V_0^{-1/4})$, so the quartic term changes these low levels by $O(1)$. Tunnelling between successive deep wells gives exponentially narrow <energy bands>. The secondary minima at $x=(2j+1)\pi$ have potential zero, and their levels lie far above the first two deep-well levels, whose energies are $-2V_0+O(\sqrt{V_0})$. Therefore the gap between the first and second bands is the first oscillator spacing to leading order:
$$
\boxed{\Delta E\sim2\sqrt{\frac{5V_0}{2}}=\sqrt{10V_0},\qquad V_0\to\infty.}
$$
The $O(1)$ anharmonic shifts and exponentially small bandwidths do not change this leading estimate.
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