MCQ
The solution of ${\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = \pm \frac{\pi }{3}$ is
- A$ \pm \frac{1}{3}$
- B$ \pm \frac{1}{4}$
- C$ \pm \frac{{\sqrt 3 }}{2}$
- ✓$ \pm \frac{1}{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}$
==> ${x^2} = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$.
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