1. The problem asks to find the exact value of $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.\n\n2. Recall that $\cos^{-1}(x)$, or arccosine, gives the angle $\theta$ in the range $[0, \pi]$ such that $\cos(\theta) = x$.\n\n3. We need to find $\theta$ where $\cos(\theta) = -\frac{\sqrt{3}}{2}$.\n\n4. From the unit circle, $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. Since cosine is negative in the second quadrant, the angle with cosine $-\frac{\sqrt{3}}{2}$ is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.\n\n5. Therefore, $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}$.\n\nFinal answer: $\boxed{\frac{5\pi}{6}}$