Question

If $\sin^{-1} \left(1-x\right)-2 \sin^{-1}x = \frac{\pi}{2}, $ then x equals

• A. $0 , - \frac{1}{2}$
• B. $0 , \frac{1}{2}$
• C. 0
• D. none of these

Answer 1

Answer: (B)

Let $\sin^{-1} \left(1-x\right)-2 \sin^{-1}x = \frac{\pi}{2}$
$ \Rightarrow \sin^{-1} \left(1-x\right) = \frac{\pi}{2}+2 \sin^{-1} x$
$ \Rightarrow \left(1-x\right) = \sin\left(\frac{\pi}{2} + 2 \sin^{-1} x\right) $
$\Rightarrow 1-x = \cos \left(2\sin^{-1} x \right)$
$ \left(\because \sin\left(90^{\circ} +\theta\right) = \cos\theta\right)$
$ \Rightarrow 1-x = \cos \left[\cos^{-1} \left(1-2 x^{2}\right)\right]$
$ \Rightarrow 1-x = 1-2 x^{2} $
$\Rightarrow 2x^{2} + 1 -x - 1 = 0 \Rightarrow 2x^{2} - x = 0 $
$\Rightarrow x\left(2x-1\right) = 0 \Rightarrow x = 0, \frac{1}{2}$