Considering the principal values of the inverse trigonometric functions, $\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right),-\frac{1}{2}< x<\frac{1}{\sqrt{2}}$, is equal to

<p>$$\begin{aligned} & \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right),-\frac{1}{2}< x<\frac{1}{\sqrt{2}} \\ & \text { Let } x=\cos \theta, \theta \in\left(\frac{\pi}{4}, \frac{2 \pi}{3}\right) \\ & \Rightarrow \sqrt{1-x^2}=\sin \theta \text { as } \sin \theta>0 \end{aligned}$$</p> <p>$\begin{aligned} \sin ^{-1}\left(\frac{\sqrt{3}}{2} \cos \theta+\frac{1}{2} \sin \theta\right)=\sin ^{-1}( & \left.\sin \left(\frac{\pi}{3}+\theta\right)\right) \\ & \frac{\pi}{3}+\theta \in\left(\frac{7 \pi}{2}, \pi\right)\end{aligned}$</p> <p>$$\begin{aligned} & =\sin ^{-1}\left(\sin \left(\pi-\left(\frac{\pi}{3}+\theta\right)\right)\right) \\ & =\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}-\theta\right)\right) \\ & =\frac{2 \pi}{3}-\theta \\ & =\frac{2 \pi}{3}-\cos ^{-1} x \\ & =\frac{2 \pi}{3}-\left(\frac{\pi}{2}-\sin ^{-1} x\right) \\ & =\frac{\pi}{6}+\sin ^{-1} x \end{aligned}$$</p>