The solution of \[{{\sin }^{-1}}x-{{\sin }^{-1}}2x=\pm \frac{\pi }{3}\] is [Karnataka CET 2005]

The solution of \[{{\sin }^{-1}}x-{{\sin }^{-1}}2x=\pm \frac{\pi }{3}\

1.	The solution of \[{{\sin }^{-1}}x-{{\sin }^{-1}}2x=\pm \frac{\pi }{3}\] is [Karnataka CET 2005]
A.	\[\pm \frac{1}{3}\]
B.	\[\pm \frac{1}{4}\]
C.	\[\pm \frac{\sqrt{3}}{2}\]
D.	\[\pm \frac{1}{2}\]
Answer» E.

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The solution of \[{{\sin }^{-1}}x-{{\sin }^{-1}}2x=\pm \frac{\pi }{3}\] is [Karnataka CET 2005]

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