If $ x \, ∈ \left[-\frac{1}{2}, 1\right]$, then $ sin^{-1}\left(\frac{\sqrt{3}}{2}x -\frac{1}{2}\sqrt{1-x^2}\right)$, equals

$ sin^{-1}x-\frac{\pi}{6}$

Let $ x = sin \theta $. Then,

$-\frac{1}{2}≤ x ≤ 1 ⇒ -\frac{1}{2}≤ \sin\theta ≤ 1 ⇒ -\frac{\pi}{6}≤ \theta ≤ \frac{\pi}{2}$

Now,

$ sin^{-1}\left(\frac{\sqrt{3}}{2}x -\frac{1}{2}\sqrt{1-x^2}\right)$

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

$ = sin^{-1} \begin{Bmatrix}sin \left(\theta -\frac{\pi}{6}\right)\end{Bmatrix}$

$= \theta - \frac{\pi}{6}$         $\left[∵ -\frac{\pi}{6}≤ \theta ≤ \frac{\pi}{2} ⇒ -\frac{\pi}{3}≤\theta -\frac{\pi}{6}≤\frac{\pi}{3}\right]$

$= sin^{-1}x-\frac{\pi}{6}$