Question

${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2},$ then $x$ is equal to

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

Answer 1

The correct option is D $0$

Given: ${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$
Let ${\sin }^{-1}x=\theta \text{so},x=\sin \theta$
$\Rightarrow {\sin }^{-1}(1-x)-2\theta =\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}(1-x)=\frac{\pi }{2}+2\theta$
$\Rightarrow 1-x=\sin (\frac{\pi }{2}+2\theta )$
$\Rightarrow 1-x=\cos 2\theta$
$\Rightarrow 1-x=1-2{\sin }^2\theta$
$\Rightarrow 1-x=1-2x^2$
$\{\because x=\sin \theta \}$
$\Rightarrow x(2x-1)=0$
$\Rightarrow x=0\text{or}2x-1=0$
$\Rightarrow x=0\text{or}x=1/2$
$x=\frac{1}{2}$ does not satisfy given equation.
So, the answer is $x=0$.
Option $\left(C\right)$ is correct.