The trignometric equation $sin^{-1}$ x = 2 $sin^{-1}$ a has a solution for

$|a| ≤ \frac{1}{\sqrt{2}}$

Let $\alpha $ be a solution of $sin^{-1}x = 2 sin^{-1}a$. Then, $sin^{-1} \alpha - 2 sin^{-1} a $

$⇒ -\frac{\pi}{2} ≤ 2sin^{-1} a ≤\frac{\pi}{2}$    $[∵-\frac{\pi}{2} ≤ sin^{-1} \alpha ≤ \frac{\pi}{2}]$

$⇒ -\frac{\pi}{4} ≤ sin^{-1} a ≤\frac{\pi}{4}$

$⇒ sin\left(-\frac{\pi}{4}\right) ≤a ≤ sin \frac{\pi}{4} ⇒ -\frac{1}{\sqrt{2}} ≤ a ≤ \frac{1}{\sqrt{2}}⇒ |a| ≤\frac{1}{\sqrt{2}}$