Solve for x :

To find: value of x

Given: $\cos \left(\sin ^{-1} x\right)=\frac{1}{2}$

$\mathrm{LHS}=\cos \left(\sin ^{-1} \mathrm{x}\right)$

$=\cos \left(\cos ^{-1}\left(\sqrt{1-\mathrm{x}^{2}}\right)\right)$

$=\sqrt{1-\mathrm{x}^{2}}$

Therefore, $\sqrt{1-\mathrm{x}^{2}}=\frac{1}{2}$

Squaring both sides,

$1-x^{2}=\frac{1}{4}$

$x^{2}=1-\frac{1}{4}$

$x^{2}=\frac{3}{4}$

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

Therefore, $x=\pm \frac{\sqrt{3}}{2}$ are the required values of $x$.