Question

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

• A. $0$
• B. $\frac{1}{2}$
• C. (a) and (b) above.
• D. $\frac{1}{4}$

Answer 1

The correct option is A $0$

${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$
${\sin }^{-1}(1-x)=(\frac{\pi }{2}-2{\sin }^{-1}x)$
$\Rightarrow 1-x=\sin \frac{\pi }{2}\cos \left(2{\sin }^{-1}x\right)-\cos \frac{\pi }{2}\sin \left(2{\sin }^{-1}x\right)$
$\Rightarrow 1-x=\cos \left(2{\sin }^{-1}x\right)$
$\Rightarrow 1-x=\cos \left(2(\frac{\pi }{2}-{\cos }^{-1}x)\right)=\cos (\pi -2{\cos }^{-1}x)$
$\Rightarrow 1-x=-\cos \left(2{\cos }^{-1}x\right)$
$\Rightarrow$ $x-1=\cos \left(2{\cos }^{-1}x\right)$
$\Rightarrow x-1=2x^2-1$
$\Rightarrow 2x^2-x=0$
$\Rightarrow x=0$ or $x=\frac{1}{2}$
$\frac{1}{2}$does not satisfy the equation.
Hence $x=0$

Alternate method:
Substituting the values, we find that only $x=0$ satisfies the given equation.