Question

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

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

Answer 1

The correct option is C $0$

We have ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}(1-x)=\frac{\pi }{2}+2{\sin }^{-1}x$
$\Rightarrow (1-x)=\sin (\frac{\pi }{2}+2{\sin }^{-1}x)\Rightarrow 1-x=\cos \left(2{\sin }^{-1}x\right)\Rightarrow 1-x=\cos \left({\cos }^{-1}(1-{2x}^2)\right)\Rightarrow 1-x=1-{2x}^2\Rightarrow {2x}^2-x=0\Rightarrow x=0,\frac{1}{2}$
But $x=\frac{1}{2}$ does not satisfy the equation,
Thus $x=0$ is the only solution.