The set of values of $x$in $(-\pi ,\pi )$ satisfying the inequation $|4\sin x-1|<\sqrt{5}$, is

We have,  $|4\sin x-1|<\sqrt{5}$

$\begin{array}{l}\Rightarrow 1-\sqrt{5}<4\sin x<1+\sqrt{5}\\ \Rightarrow -\frac{\sqrt{5}-1}{4}<\sin x<\frac{\sqrt{5}+1}{4}\\ \Rightarrow -\sin \frac{\pi }{10}<\sin x<\cos \frac{\pi }{10}\\ \Rightarrow \sin \left(-\frac{\pi }{10}\right)<\sin x<\sin \left(\frac{\pi }{2}-\frac{\pi }{10}\right)\\ \Rightarrow \sin \left(-\frac{\pi }{10}\right)<\sin x<\sin \frac{3\pi }{10}\end{array}$

$\Rightarrow -\frac{\pi }{10}<x<\frac{3\pi }{10}\left[\because \sin x\text{ is increasing on }\left(-\frac{\pi }{2},\frac{\pi }{2}\right)\right]$

$\Rightarrow x\in \left(-\frac{\pi }{10},\frac{3\pi }{10}\right)$