A) \[\pm \frac{1}{3}\]
B) \[\pm \frac{1}{4}\]
C) \[\pm \frac{\sqrt{3}}{2}\]
D) \[\pm \frac{1}{2}\]
Correct Answer: D
Solution :
\[{{\sin }^{-1}}2x={{\sin }^{-1}}x-{{\sin }^{-1}}\frac{\sqrt{3}}{2}\] \[{{\sin }^{-1}}2x={{\sin }^{-1}}\left( x\sqrt{\left( 1-\frac{3}{4} \right)}-\frac{\sqrt{3}}{2}\sqrt{1-{{x}^{2}}} \right)\] \[2x=\left( \frac{x}{2}-\frac{\sqrt{3}}{2}\sqrt{1-{{x}^{2}}} \right)\] \[\frac{\sqrt{3}}{2}\sqrt{1-{{x}^{2}}}=\frac{x}{2}-2x=\frac{-3x}{2}\] \[\frac{3(1-{{x}^{2}})}{4}=\frac{9{{x}^{2}}}{4}\] \[\Rightarrow 3-3{{x}^{2}}=9{{x}^{2}}\]Þ \[{{x}^{2}}=\frac{1}{4}\Rightarrow x=\pm \frac{1}{2}\].
You need to login to perform this action.
You will be redirected in
3 sec
