First, we simplify the argument of the inverse cosine function:

$\cos(-\frac{7\pi}{3}) = \cos(\frac{7\pi}{3})$ (Since cosine is an even function, cos(-x) = cos(x))

$\cos(\frac{7\pi}{3}) = \cos(2\pi + \frac{\pi}{3}) = \cos(\frac{\pi}{3})$ (Since cosine has a period of $2\pi$)

$\cos(\frac{\pi}{3}) = \frac{1}{2}$

Now, we evaluate the inverse cosine:

$\cos^{-1}[\cos(-\frac{7\pi}{3})] = \cos^{-1}[\frac{1}{2}]$

Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, we have $\cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$