We have
$\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
Let $A=\sin ^{-1} x$
$\begin{array}{ll}
\Rightarrow & x=\sin A, \cos A=\sqrt{1-x^2} \\
\text {and } & B=\cos ^{-1} x \\
\Rightarrow & x=\cos B, \sin B=\sqrt{1-x^2}
\end{array}$
Now,
$\begin{aligned}
\sin (A-B) & =\sin A \cos B-\cos A \sin B \\
& =x \cdot x-\sqrt{1-x^2} \cdot \sqrt{1-x^2} \\
& =x^2-1+x^2 \\
\sin \left(\frac{\pi}{6}\right) & =2 x^2-1 \\
\Rightarrow 2 x^2-1 & =\frac{1}{2} \Rightarrow 2 x^2=\frac{3}{2} \Rightarrow x^2=\frac{3}{4} \\
\Rightarrow \quad x & =\frac{\sqrt{3}}{2}
\end{aligned}$