Question

The set of all x in $(\frac{\pi }{2},\frac{-\pi }{2})$ satisfying $\left|4\sin x-1\right|<\sqrt{5}$

Answer 1

We have,

$f\left(x\right)=|4\sin x-1|<\sqrt{5}$

If $x\in \frac{\pi }{2}$

Then,

$f\left(x\right)=|4\sin x-1|$

$f\left(\frac{\pi }{2}\right)=|4\sin \frac{\pi }{2}-1|$

$f\left(\frac{\pi }{2}\right)=|4\times 1-1|$

$f\left(\frac{\pi }{2}\right)=\left|3\right|$

$f\left(\frac{\pi }{2}\right)=3>\sqrt{5}$

If $x\in -\frac{\pi }{2}$

Then,

$f\left(x\right)=|4\sin x-1|$

$f\left(\frac{\pi }{2}\right)=|4\sin (-\frac{\pi }{2})-1|$

$f\left(\frac{\pi }{2}\right)=|4\times (-1)-1|$

$f\left(\frac{\pi }{2}\right)=|-5|$

$f\left(\frac{\pi }{2}\right)=5>\sqrt{5}$

Hence, this is the answer.