$2{\cos ^{ - 1}}x = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)$
અહી ${\cos ^{ - 1}} - x = \theta ,\theta \in [o,\pi ]$
$\therefore x = \cos \theta$
$\therefore 2\theta = {\sin ^{ - 1}}\left( {2\cos \theta \sqrt {1 - {{\cos }^2}\theta } } \right)$
$\therefore 2 \theta = \sin^{-1} (2 \cos \theta \sin \theta)$
$\therefore 2 \theta = \sin^{-1} (\sin2 \theta)$
$\therefore 2 \theta = 2 \theta$
હવે $-\frac{\pi}{2} \leq 2 \theta \leq \frac{\pi}{2}$
$\therefore -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$ તથા $\theta \in [0,\pi ]$
હોવાથી $\theta \in \left[ {o,\frac{\pi }{4}} \right]$
પ્રથમ ચરણમાં $\cos$ ઘટતું વિધેય છે.
$\therefore 0 \leq \theta \leq \frac{\pi}{4} \Rightarrow \cos 0 \geq \cos \theta \geq \cos \frac{\pi}{4}$
$\therefore \frac{1}{{\sqrt 2 }} \leqslant \cos \theta \leqslant 1$
$\therefore x \in \left[\frac{1}{\sqrt{2}}, 1\right]$
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