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Question:

If $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$, then the value of $x$ is ______________.

Solution:

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

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

$\Rightarrow \cos 2 \theta=\frac{1}{9}$

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

$\Rightarrow 2 \sin ^{2} \theta=1-\frac{1}{9}=\frac{8}{9}$

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

$\Rightarrow \sin \theta=\pm \frac{2}{3}$

$\Rightarrow x=\pm \frac{2}{3}$                            $(\sin \theta=x)$

Thus, the value of $x$ is $\pm \frac{2}{3}$.

If $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$, then the value of $x$ is $\pm \frac{2}{3}$

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