Answer :
Answer:
[tex]\large\boxed{x=-\dfrac{\pi}{3}+k\pi\ \vee\ x=\dfrac{\pi}{3}+k\pi}\\\vee\\\boxed{x=-\dfrac{\pi}{4}+k\pi\ \vee\ x=\dfrac{\pi}{4}+k\pi}\\ k\in\mathbb{Z}[/tex]
Step-by-step explanation:
[tex]\tan^4x-4\tan^2x+3=0\\\\Domain:\ x\neq\dfrac{\pi}{2}+k\pi,\ k\in\mathbb{Z}\\\\\text{Substitution}\ \tan^2x=t\geq0\\\\\text{Therefore we have the equation:}\\\\t^2-4t+3=0\\\\t^2-3t-t+3=0\\\\t(t-3)-1(t-3)=0\\\\(t-3)(t-1)=0\iff t-3=0\ \vee\ t-1=0\\\\t-3=0\qquad\text{add 3 to both sides}\\\boxed{t=3}\\\\t-1=0\qquad\text{add 1 to both sides}\\\boxed{t=1}[/tex]
[tex]\text{We're going back to substitution}\\\\t=3\to\tan^2x=3\to\tan x=\pm\sqrt3\\\\x=-\dfrac{\pi}{3}+k\pi\ \vee\ x=\dfrac{\pi}{3}+k\pi,\ k\in\mathbb{Z}\\\\t=1\to\tan^2x=1\to\tan x=\pm1\\\\x=-\dfrac{\pi}{4}+k\pi\ \vee\ x=\dfrac{\pi}{4}+k\pi,\ k\in\mathbb{Z}[/tex]