Question

`\tan^(4) x - 4 \tan^(2) + 3 = 0`
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Answer 1

`\large\boxed{x=-(\pi)/(3)+k\pi\ \vee\ x=(\pi)/(3)+k\pi}\\\vee\\\boxed{x=-(\pi)/(4)+k\pi\ \vee\ x=(\pi)/(4)+k\pi}\\ k\in\mathbb{Z}`

Explanation:

`\tan^4x-4\tan^2x+3=0\\\\Domain:\ x\\eq(\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}`

`\text{We're going back to substitution}\\\\t=3\to\tan^2x=3\to\tan x=\pm\sqrt3\\\\x=-(\pi)/(3)+k\pi\ \vee\ x=(\pi)/(3)+k\pi,\ k\in\mathbb{Z}\\\\t=1\to\tan^2x=1\to\tan x=\pm1\\\\x=-(\pi)/(4)+k\pi\ \vee\ x=(\pi)/(4)+k\pi,\ k\in\mathbb{Z}`