# How can this be solved?

## $3 {\tan}^{2} \left(x\right) + 13 \tan \left(x\right) - 10$

Aug 1, 2018

Note:To complete equation we have to take,
$3 {\tan}^{2} x + 13 \tan x - 10 = 0$

$x = \left\{k \pi + \arctan \left(\frac{2}{3}\right) , k \in \mathbb{Z}\right\} \cup \left\{k \pi - \arctan \left(5\right) , k \in \mathbb{Z}\right\}$

#### Explanation:

Here ,

$3 {\tan}^{2} x + 13 \tan x - 10 = 0$

$\implies 3 {\tan}^{2} x - 2 \tan x + 15 \tan x - 10 = 0$

$\implies \tan x \left(3 \tan x - 2\right) + 5 \left(3 \tan x - 2\right) = 0$

$\implies \left(3 \tan x - 2\right) \left(\tan x + 5\right) = 0$

$\implies 3 \tan x - 2 = 0 \mathmr{and} \tan x + 5 = 0$

$\implies \tan x = \frac{2}{3} \mathmr{and} \tan x = - 5$

$\left(i\right) \tan x = \frac{2}{3} \implies x = k \pi + \arctan \left(\frac{2}{3}\right) , k \in \mathbb{Z}$

$\left(i i\right) \tan x = - 5 \implies x = k \pi + \arctan \left(- 5\right) , k \in \mathbb{Z}$

$i . e . x = k \pi - \arctan \left(5\right) , k \in \mathbb{Z}$

Note:To complete equation we have to take,

$3 {\tan}^{2} x + 13 \tan x - 10 = 0$

Aug 1, 2018

$\tan x = \frac{2}{3}$

$x = 0.588 + n \pi$ where $n$ is an integer

$\tan x = - 5$

$x = 1.77 + n \pi$ where $n$ is an integer

#### Explanation:

$3 {\tan}^{2} x + 13 \tan x - 10$

Think of $\tan x$ as $y$

So, $3 {\tan}^{2} x + 13 \tan x - 10$ becomes $3 {y}^{2} + 13 y - 10$

Then, using the quadratic formula,

$y = \frac{- 13 \pm \sqrt{{13}^{2} - 4 \left(3\right) \left(- 10\right)}}{2 \times 3}$

$y = \frac{- 13 \pm \sqrt{289}}{6}$

$y = \frac{- 13 \pm 17}{6}$

$y = \frac{- 13 + 17}{6}$ and $y = \frac{- 13 - 17}{6}$

$y = \frac{2}{3}$ or $y = - 5$

Since $y = \tan x$, then

$\tan x = \frac{2}{3}$ or $\tan x = - 5$

$\tan x = \frac{2}{3}$

$x = 0.588 + n \pi$ where $n$ is an integer

$\tan x = - 5$

$x = 1.77 + n \pi$ where $n$ is an integer