For a particle on a ring (with radius $R$ and changing angle $\theta$) with only kinetic energy ($V=0$) we get the expressions for the wavefunction (normalized) and eigenvalues $$\Psi_n (\theta) = \frac{1}{\sqrt{2 \pi}} e^{in \theta}$$ $$E_n= \frac{\hbar ^2}{2mR^2}n^2$$ with $$n=0, \pm1, \pm2, ...$$
This means we have degeneracy for $|n|>0$, and only one eigenstate for $n=0$ with energy $E_0 = 0$.
Now using the WKB method, considering $L=2\pi R$ the length of the ring and $s$ the distance travelled by the particle, for $E > V(s)$, being $V(s)$ any potential we get $$\Psi(s) = \frac{1}{\sqrt{p(s)}}(C_+ e^{i \phi (s)} + C_- e^{-i \phi (s)}) \tag{1. a}$$ or similarly $$\Psi(s) = \frac{1}{\sqrt{p(s)}}(A\, cos(\phi (s)) + B\, sin(\phi (s))) \tag{1. b}$$ with $$\phi(s) = \int_0^s p(s') ds'$$ and $$p(s)=\sqrt{2m(E_n - V(s))}$$
After the boundary conditions $\psi(s) = \psi (s + L)$, or equivalentely $\psi(0) = \psi (L)$, the quantization for the eigenvalues is obtained $$\int_0^L p(s') ds' = n h \tag{2}$$ since $p(s)>0$, we obtain $n=1, 2, ...$ by evaluation of ($2$).
Physically speaking we could think of a particle moving in the counter clockwise direction ($s=0 \rightarrow s=L$) regarding ($2$). Inverting direction and the eigenvalues $E$ would remain the same, but now we have $n=-1, -2, ...$ There's degeneracy like the example above with $n=\pm 1, \pm 2, ...$ however $n=0$ doesn't exist here (for it regards to $p=0$ and the WKB approximation doesn't apply). The degeneracy is sustained by the fact that in ($1$) there exists two linearly independent solutions.
My question is: How can I obtain the wavefunction ($1.b$) for each value $E_n$? Either obtaining the constants ($A$ and $B$) or a more elegant expression.
