MCQ
If $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$, then $x=$
  • A
    Only $\frac{2}{3}$
  • B
    Only $\frac{-2}{3}$
  • $\frac{2}{3}, \frac{-2}{3}$
  • D
    Neither $\frac{2}{3}$ nor $\frac{-2}{3}$

Answer

Correct option: C.
$\frac{2}{3}, \frac{-2}{3}$
(C) Let $\sin ^{-1} x=\theta \Rightarrow x=\sin \theta$
$\cos \left(2 \sin ^{-1} x\right)-\frac{1}{9} \quad \Rightarrow \cos 2 \theta-\frac{1}{9}$
$\Rightarrow 1-2 \sin ^2 \theta=\frac{1}{9} \quad \Rightarrow 1-2 x^2=\frac{1}{9}$
$\Rightarrow 2 x^2=1-\frac{1}{9}=\frac{8}{9} \quad \Rightarrow x^2=\frac{4}{9}$
$\Rightarrow x= \pm \frac{2}{3}$

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