inv-trig

Question: The solution of ${{\sin }^{-1}}x-{{\sin }^{-1}}2x=\pm \frac{\pi }{3}$ is



1) $\pm \frac{1}{3}$
2) $\pm \frac{1}{4}$
3) $\pm \frac{\sqrt{3}}{2}$
4) $\pm \frac{1}{2}$
Solution: Explanation: No Explanation
Inverse trigonometric functions

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