Question

If set of all values of $x\in (-\frac{\pi }{2},\frac{\pi }{2})$ satisfying $|4\sin x+\sqrt{2}|<\sqrt{6}$ is $(\frac{a\pi }{24},\frac{b\pi }{24})$, then the value of $\left|\frac{a-b}{3}\right|=$

Answer 1

$|4\sin x+\sqrt{2}|<\sqrt{6}$
$\Rightarrow -\sqrt{6}<4\sin x+\sqrt{2}<\sqrt{6}\Rightarrow -\sqrt{2}(\sqrt{3}+1)<4\sin x<\sqrt{2}(\sqrt{3}-1)$
$\Rightarrow -\frac{(\sqrt{3}+1)}{2\sqrt{2}}<\sin x<\frac{(\sqrt{3}-1)}{2\sqrt{2}}\Rightarrow -\sin \frac{5\pi }{12}<\sin x<\sin \frac{\pi }{12}\Rightarrow -\frac{5\pi }{12}<x<\frac{\pi }{12}\text{for}x\in (-\frac{\pi }{2},\frac{\pi }{2})$
Comparing with $\frac{a\pi }{24}<x<\frac{b\pi }{24},$ we get, $a=-10,b=2$
$\therefore \left|\frac{a-b}{3}\right|=\left|\frac{-10-2}{3}\right|=4$