Operating modulus on inverse trigonometric inequality

Question

So, I was going through this example of an inverse trigonometric inequality in my Math textbook and I've come up to this step:

$$-2\pi < 3\tan^{-1}x-\pi /2< \pi $$ Now in the next step they have operated modulus on the inverse trigonometric function like this: $$ 0 \leq |3\tan^{-1}x-\pi /2|< 2\pi $$ I am not clear as to how the LHS and RHS inequality, i.e. 0 and 2, respectively, have arrived by operating modulus on the inverse trigonometric function as they seem to have skipped that step in the example. Could someone please help me out?

Answer 1

The range of $\arctan$ is $[\frac {-\pi} 2,\frac{\pi}2]$. Triple that gets you $[\frac {-3\pi} 2,\frac{3\pi}2]$ as a range. Finally subtract $\frac \pi 2$ from each number and you get a range of $[-2\pi,\pi]$, which gets you that the max distance from the origin is $2\pi$ (and minimum is 0)

Answer 2

If $-2\pi < 3\tan^{-1}x-\pi /2< \pi,$ then $-2\pi < 3\tan^{-1}x-\pi /2< 2\pi.$

If $-2\pi<a<2\pi$ then $|a|<2\pi.$

And the $0$ comes from the fact that all absolute values are $\ge0.$