# How do you solve \tan ^ { 2} x - 4\tan x + 1= 0?

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Feb 9, 2018

$\textcolor{b l u e}{\frac{5 \pi}{12} + n \pi}$

$\textcolor{b l u e}{\frac{\pi}{12} + n \pi}$

#### Explanation:

${\tan}^{2} x - 4 \tan x + 1 = 0$

Let $\textcolor{w h i t e}{88} u = \tan x$

Then:

${u}^{2} - 4 u + 1 = 0$

$u = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - \left(4 \cdot 1 \cdot 1\right)}}{2 \cdot 1}$

$u = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$

$u = \tan x = 2 \pm \sqrt{3}$

$x = \arctan \left(2 + \sqrt{3}\right) = \textcolor{b l u e}{\frac{5 \pi}{12} + n \pi}$

$x = \arctan \left(2 - \sqrt{3}\right) = \textcolor{b l u e}{\frac{\pi}{12} + n \pi}$

For: $\left\{n \in \mathbb{Z}\right\}$

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