A) \[\pm \frac{1}{3}\]
B) \[\pm \frac{1}{3}\]
C) \[\pm \frac{\sqrt{3}}{2}\]
D) \[\pm \frac{1}{2}\]
Correct Answer: D
Solution :
\[{{\sin }^{-1}}x-{{\sin }^{-1}}2x\pm \frac{\pi }{3}\] \[\Rightarrow \,\,{{\sin }^{-1}}x-{{\sin }^{-1}}\left( \pm \frac{\sqrt{3}}{2} \right)={{\sin }^{-1}}2x\] \[\Rightarrow \,\,{{\sin }^{-1}}\left[ x\sqrt{1-\frac{3}{4}}-\left( \pm \frac{\sqrt{3}}{2}\sqrt{1-{{x}^{2}}} \right) \right]={{\sin }^{-1}}\] \[\Rightarrow \] \[\frac{x}{2}-\left( \pm \frac{\sqrt{3}}{2}\sqrt{1-{{x}^{2}}} \right)=2x\] \[\Rightarrow \] \[-\left( \pm \sqrt{3}\sqrt{1-{{x}^{2}}} \right)=3x\] On squaring, both sides we get, \[3\,(1-{{x}^{2}})=9{{x}^{2}}\] \[\Rightarrow \] \[1-{{x}^{2}}=3{{x}^{2}}\] \[\Rightarrow \] \[4{{x}^{2}}=1\] \[\Rightarrow \] \[x=\pm \frac{1}{2}\]
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