Question

The value(s) of $x$ satisfying the equation ${\left(sin\right)}^{-1}\left(1-x\right)-2{\left(sin\right)}^{-1}x=\frac{\pi }{2}$ is/are

• A. $0$
• B. $\frac{1}{2}$
• C. $0,\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A)

We have, ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}(1-x)=\frac{\pi }{2}+2{\sin }^{-1}x$
$\Rightarrow 1-x=\sin \left(\frac{\pi }{2}+2{\sin }^{-1}x\right)$
$\Rightarrow 1-x=\cos \left(2{\sin }^{-1}x\right)$
$\Rightarrow 1-x=\cos \left\{{\cos }^{-1}\left(1-2x^2\right)\right\}\left[\because 2{\sin }^{-1}x={\cos }^{-1}\left(1-2x^2\right)\right]$
$\Rightarrow 1-x=\left(1-2x^2\right)$
$\Rightarrow x=2x^2\Rightarrow x(2x-1)=0\Rightarrow x=0,\frac{1}{2}$
For, $x=\frac{1}{2},$ we have $LHS={\sin }^{-1}(1-x)-2{\sin }^{-1}x$
$={\sin }^{-1}\frac{1}{2}-2{\sin }^{-1}\frac{1}{2}=-{\sin }^{-1}\frac{1}{2}=\frac{-\pi }{6}\ne R.H.S.$
So, $x=\frac{1}{2}$ is not a root of the given equation. Clearly, $x=0$ satisfies the equation

Here, $x=0$ is the root of the given equation.