$2\cos^{-1}x = \sin^{-1}(2x\sqrt{1-x^2})$ is valid for all values of $x$ satisfying

Question

$-1 \leq x \leq 1$

$0 \leq x \leq 1$

$\frac{1}{\sqrt{2}} \leq x \leq 1$

$0 \leq x \leq \frac{1}{\sqrt{2}}$

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