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

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

Solution:

To find: value of $x$
Formula Used: $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ .
Given: $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$
$\text { LHS }=\cos \left(2 \sin ^{-1} \mathrm{x}\right)$
Let $\theta=\sin ^{-1} x$
So, $x = sin \theta \dots (1)$
$\begin{array}{l} \text { LHS }=\cos (2 \theta) \\ =1-2 \sin ^{2} \theta \end{array}$
Substituting in the given equation,
$\begin{array}{l} 1-2 \sin ^{2} \theta=\frac{1}{9} \\ 2 \sin ^{2} \theta=\frac{8}{9} \\ \sin ^{2} \theta=\frac{4}{9} \end{array}$
Substituting in (1),
$\begin{array}{l} x^{2}=\frac{4}{9} \\ x=\pm \frac{2}{3} \end{array}$
Therefore, $x=\pm \frac{2}{3}$ are the required values of $x$ .