Solving \[2{{\cos }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}}),\]we get

\[x\in \left[ \frac{\sqrt{2}}{2},1 \right]\]

Solving \[2{{\cos }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}}),\]we g

1.	Solving \[2{{\cos }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}}),\]we get
A.	\[x\in \left[ \frac{\sqrt{2}}{2},1 \right]\]
B.	\[x=3\]
C.	\[x\in [3,4]\]
D.	\[x=0\]
Answer» B. \[x=3\]

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Solving \[2{{\cos }^{-1}}x={{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}}),\]we get

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