TXD:D=R\{${\frac{\pi }{2}+k\pi}$}
giả sử có 1 giá trị P sao cho $x\in D$
Ta có P=$\frac{2(\tan x)^{2}+12\tan x}{(\tan x)^{2}+2\tan x+3}$
<=>(P-2)$(\tan x)^{2}$ +(2P-12)$\tan x$+3P=0
$\Delta $=$(P-6)^{2}$-3P(P-2)=$-2P^{2}$-6P+36$\geq $0
<=>$-6\leq P\leq 3$
MaxP=3<=>$x=\arctan(3)+k\pi$
MinP=-6<=>$x=\arctan (-\frac{3}{2})+k\pi $