Solve for x:

To find: value of x

Formula Used: $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$

Given: $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$

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

Let $\theta=\sin ^{-1} x$

So, $x=\sin \theta \ldots$ (1)

$\mathrm{LHS}=\cos (2 \theta)$

$=1-2 \sin ^{2} \theta$

Substituting in the given equation,

$1-2 \sin ^{2} \theta=\frac{1}{9}$

$2 \sin ^{2} \theta=\frac{8}{9}$

$\sin ^{2} \theta=\frac{4}{9}$

Substituting in (1),

$x^{2}=\frac{4}{9}$

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

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