$\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ equals:

The inverse cosine function $\cos^{-1}(x)$ has range $[0, \pi]$. When evaluating $\cos^{-1}(\cos\theta)$, the result is the angle in $[0, \pi]$ that has the same cosine value as $\theta$.

Finding the value of $\cos\frac{7\pi}{6}$:

$\frac{7\pi}{6} = \pi + \frac{\pi}{6}$

This angle is in the third quadrant where cosine is negative.

Using the reference angle $\frac{\pi}{6}$:

$\cos\frac{7\pi}{6} = -\cos\frac{\pi}{6}$

$\cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$

Finding which angle in $[0, \pi]$ has cosine equal to $-\frac{\sqrt{3}}{2}$:

Since cosine is negative, the angle must be in the second quadrant, between $\frac{\pi}{2}$ and $\pi$.

The angle with reference angle $\frac{\pi}{6}$ in the second quadrant:

$\pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6}$

$= \frac{5\pi}{6}$

Checking: $\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$

Since $\frac{5\pi}{6} \in [0, \pi]$:

$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$