# How do I solve, tan^2(x)-(1+sqrt(3))tanx+sqrt(3)<0 ?

## I know how to factor it, but then how would I get to the solutions? The domain is [pi/2,5pi/2].

Apr 11, 2018

color(blue)((pi/4,pi/3)uu((5pi)/4,(4pi)/3)uu((9pi)/4,(7pi)/3)

#### Explanation:

${\tan}^{2} \left(x\right) - \left(1 + \sqrt{3}\right) \tan \left(x\right) + \sqrt{3} < 0$

Let $\setminus \setminus \setminus \setminus \setminus u = \tan \left(x\right)$

${u}^{2} - \left(1 + \sqrt{3}\right) u + \sqrt{3} < 0$

Factor LHS:

$\left(1 - u\right) \left(- u + \sqrt{3}\right) < 0$

Solving for zero to get the bounds:

$1 - u = 0 \implies u = 1$

$- u + \sqrt{3} = 0 \implies u = \sqrt{3}$

But $\setminus \setminus \setminus \setminus \setminus u = \tan \left(x\right)$

$\tan \left(x\right) = 1$

$\tan \left(x\right) = \sqrt{3}$

$x = \arctan \left(\tan \left(x\right)\right) = \arctan \left(1\right) \implies x = \frac{\pi}{4} , \frac{5 \pi}{4} , \frac{9 \pi}{4}$

$x = \arctan \left(\tan \left(x\right)\right) = \arctan \left(\sqrt{3}\right) \implies x = \frac{\pi}{3} , \frac{4 \pi}{3} , \frac{7 \pi}{3}$

Using:

$\left(1 - u\right) \left(- u + \sqrt{3}\right) < 0$

and substituting $u = \tan \left(x\right)$

$\left(1 - \tan \left(x\right)\right) \left(- \tan \left(x\right) + \sqrt{3}\right) < 0$

For:

$\frac{\pi}{4} < x < \frac{\pi}{3} = 3$

(1-tan((5pi)/18))(-tan((5pi)/18))+sqrt(3))<0 \ \ \  true

For:

$\frac{5 \pi}{4} < x < \frac{4 \pi}{3}$

(1-tan((23pi)/18))(-tan((23pi)/18))+sqrt(3))<0 \ \ \  true

For:

$\frac{9 \pi}{4} < x < \frac{7 \pi}{3}$

(1-tan((55pi)/24))(-tan((55pi)/24))+sqrt(3))<0 \ \ \  true

Solutions:

color(blue)((pi/4,pi/3)uu((5pi)/4,(4pi)/3)uu((9pi)/4,(7pi)/3)