Question

The solution of equation
${\sin }^{-1}(1-\frac{x}{2})-2{\sin }^{-1}\left(\frac{x}{2}\right)=\frac{\pi }{2}$ is

• A. $x=0$
• B. $x=\frac{1}{2}$
• C. $x=0$ and $x=\frac{1}{2}$
• D. None of these

Answer 1

The correct option is A $x=0$

Given ${\sin }^{-1}(1-\frac{x}{2})-2{\sin }^{-1}\left(\frac{x}{2}\right)=\frac{\pi }{2}$
Put $x=0$, $LHS={\sin }^{-1}\left(1\right)-2{\sin }^{-1}\left(0\right)=\frac{\pi }{2}=RHS$
Put $x=\frac{1}{2}$
$LHS={\sin }^{-1}\left(\frac{3}{4}\right)-2{\sin }^{-1}\left(\frac{1}{4}\right)$
$={\sin }^{-1}\left(\frac{3}{4}\right)-{\sin }^{-1}(2\times \frac{1}{4}\sqrt{1-\frac{1}{16}})$
$={\sin }^{-1}[\frac{3}{4}\sqrt{1-\frac{15}{64}}-\frac{\sqrt{15}}{8}\times \sqrt{1-\frac{9}{16}}]={\sin }^{-1}[\frac{21}{32}-\frac{\sqrt{105}}{32}]\ne {\sin }^{-1}\left(1\right)$
Hence the required solution is $x=0$