Answer
$\dfrac{-\pi}{3}$
Work Step by Step
The range of $\sin^{-1}{x}$ is $[-\dfrac{ \pi}{2}, \dfrac{\pi}{2}]$.
Here, $\cos(\dfrac{-7\pi}{6})=\dfrac{ -\sqrt 3}{2}$, because $\sin{(\dfrac{-\pi}{3})}=\dfrac{-\sqrt 3}{2}$ and $\dfrac{-\pi}{3}$ lies in the range of $\sin^{-1}{x}$ .
Therefore, we have:
$\sin^{-1}(\cos {(\dfrac{-7\pi }{6})}]=\sin^{-1} ({\dfrac{- \sqrt 3}{2}})=\dfrac{-\pi}{3}$
