Given the finite support of the domain, and the fact that the *most* dispersed distribution in this case is certainly the uniform distribution $U$ (with $u(x) = 1/2\pi$), you could measure the dispersion via the Kullback-Leibler divergence from uniformity, which in this case is closely related to the entropy of the distribution:

$$
D_{KL}(P \mid \mid U) = - \int_0^{2\pi} p(x) \log \left( \frac{p(x)}{1/2\pi} \right) \, dx = \log \left(\frac{1}{2 \pi} \right) - \int_0^{2\pi} p(x) \log p(x) \, dx
$$

One reason to suppose that this could be preferable to standard deviation (or the c.v., which does have a problematic zero divisor case) is that this is invariant under rotations of the plane in which $x$ measures angles -- the value $x = 0 \, \textrm{rad}$ is usually an arbitrary marker point in cases where circular distributions would be use, and any other point on the circle could have been chosen as $x' = 0 \, \textrm{rad}$. Because the definition only involves a total integral and the values of $p(x)$, the result is the same no matter what reference direction is chosen as 0 radians.

The standard deviation of a distribution on on the real line is useful exactly because is invariant to linear translations of the axis, but fails to respect the underlying symmetry for the circular domain (think, for instance, if the probability spiked at a value just greater than $0$ and another value at just less than $2\pi$, the standard deviation would be large, but $0$ and $2\pi$ are close in angular space -- with a different choice of reference direction these points could have numerically similar angles, and the distribution would have small standard deviation.