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