[tex]\tan x-5=0[/tex]
[tex]\tan x=5[/tex]
Now before taking the inverse tangent of both sides, recall that
[tex]\tan x=\tan(x\pm\pi)=\tan(x\pm2\pi)=\cdots=\tan(x\pm n\pi)[/tex]
where [tex]n[/tex] is any integer. This means that when taking the inverse tangent of [tex]\tan x[/tex], instead of just [tex]x[/tex], we would end up with
[tex]\arctan(\tan x)=\arctan(\tan(x+n\pi))=x\pm n\pi[/tex]
So after doing so in the equation above, we're left with
[tex]x\pm n\pi=\arctan5\implies x=\arctan5\pm n\pi[/tex]
As [tex]n[/tex] can be any integer, including negative integers, we can simply write
[tex]x=\arctan5+n\pi,\,n\in\mathbb Z[/tex]
