If [tex]-\dfrac\pi2<x<\dfrac\pi2[/tex] (which is the domain of [tex]\tan x[/tex]), then [tex]\tan x=-1[/tex] for [tex]x=\tan^{-1}(-1)=-\tan^{-1}1=-\dfrac\pi4[/tex].
But also, recall that [tex]\tan x[/tex] has period [tex]\pi[/tex], which means [tex]\tan(x+n\pi)=\tan x[/tex] for all integers [tex]n[/tex]. This means the general solution to [tex]\tan x=-1[/tex] is [tex]-\dfrac\pi4+n\pi[/tex] for [tex]n\in\mathbb Z[/tex].
This doesn't match any of the given choices, but we can simply write
[tex]x=-\dfrac\pi4+n\pi=-\dfrac\pi4+\pi+(n-1)\pi=\dfrac{3\pi}4+(n-1)\pi[/tex]
and we can just replace [tex]n-1[/tex] with [tex]n[/tex], since both can be arbitrary integers, which means (D) is the correct answer.
