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