Question

Solve $2co{s}^{-1}x=si{n}^{-1}\left(2x\sqrt{1-x^2}\right)$.

• A. $xϵ[1/\sqrt{1},1]$
• B. $\Rightarrow x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$
• C. $\Rightarrow x\in [\frac{-2}{\sqrt{2}},\frac{1}{\sqrt{2}}]$
• D. $\Rightarrow x\in [\frac{-3}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

Answer 1

The correct option is B $\Rightarrow x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

If $\frac{-1}{\sqrt{2}}\le \theta \le \frac{1}{\sqrt{2}}$ then $2{\sin }^{-1}\theta ={\sin }^{-1}\left(2\theta \sqrt{1-{\theta }^2}\right)$
$\therefore 2{\cos }^{-1}x={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)\Rightarrow 2{\cos }^{-1}x=2{\sin }^{-1}x$
$\Rightarrow x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$