Solve for x :

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

We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$

So, $\sin ^{-1} \mathrm{x}=\frac{\pi}{2}-\cos ^{-1} \mathrm{X}$

Substituting in the given equation,

$\frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$

Rearranging,

$2 \cos ^{-1} x=\frac{\pi}{2}-\frac{\pi}{6}$

$2 \cos ^{-1} x=\frac{\pi}{3}$

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

$x=\frac{\sqrt{3}}{2}$

Therefore, $x=\frac{\sqrt{3}}{2}$ is the required value of $x$.