# How do you calculate cos^-1(sqrt3/2)?

$x = \frac{\pi}{6}$
${\cos}^{-} 1 \left(\frac{\sqrt{3}}{2}\right) = x \Leftrightarrow \cos x = \frac{\sqrt{3}}{2}$, from the definition of the inverse function.
Now, we know for $\cos x = \frac{\sqrt{3}}{2} , x = \frac{\pi}{6} + 2 n \pi , x = \frac{11 \pi}{6} + 2 n \pi$ are valid solutions; however, since the domain of the arccosine is $\left[- 1 , 1\right] , x = \frac{\pi}{6}$ is the only solution.