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The trigonometric equation $\sin ^{-1} x=2 \sin ^{-1} 2 a$ has a real solution, if
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Verified Answer
The correct answer is:
$|a| \leq \frac{1}{2 \sqrt{2}}$
We know that, $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$\Rightarrow \frac{-\pi}{2} \leq 2 \sin ^{-1} 2 a \leq \frac{\pi}{2}$
$\Rightarrow \frac{-\pi}{4} \leq \sin ^{-1} 2 a \leq \frac{\pi}{4}$
$\Rightarrow \sin \left(\frac{-\pi}{4}\right) \leq 2 a \leq \sin \left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{-1}{\sqrt{2}} \leq 2 a \leq \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{-1}{2 \sqrt{2}} \leq a \leq \frac{1}{2 \sqrt{2}} \Rightarrow|a| \leq \frac{1}{2 \sqrt{2}}$
$\Rightarrow \frac{-\pi}{2} \leq 2 \sin ^{-1} 2 a \leq \frac{\pi}{2}$
$\Rightarrow \frac{-\pi}{4} \leq \sin ^{-1} 2 a \leq \frac{\pi}{4}$
$\Rightarrow \sin \left(\frac{-\pi}{4}\right) \leq 2 a \leq \sin \left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{-1}{\sqrt{2}} \leq 2 a \leq \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{-1}{2 \sqrt{2}} \leq a \leq \frac{1}{2 \sqrt{2}} \Rightarrow|a| \leq \frac{1}{2 \sqrt{2}}$
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