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# Solve for x :

Question:

Solve for x :

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

Solution:

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 $x$.