# How do you find the x intercepts of y=tan^2((pix)/6)-3?

##### 1 Answer
Mar 18, 2017

$x = \left\{\ldots \ldots \ldots - 16 , - 14 , - 10 , - 8 , - 4 , - 2 , 2 , 4 , 8 , 10 , 14 , 16 , \ldots \ldots \ldots\right\}$

#### Explanation:

$y = {\tan}^{2} \left(\frac{\pi x}{6}\right) - 3$

$x$-intercept is when $y = 0$

$0 = {\tan}^{2} \left(\frac{\pi x}{6}\right) - 3$

or $3 = {\tan}^{2} \left(\frac{\pi x}{6}\right)$

$\tan \left(\frac{\pi x}{6}\right) = \pm \sqrt{3}$

${\tan}^{-} 1 \left(\pm \sqrt{3}\right) = \frac{\pi x}{6}$ and as $\tan \left(\pm \frac{\pi}{3}\right) = \pm \sqrt{3}$

$\frac{\pi x}{6} = n \pi \pm \frac{\pi}{3}$,, where $n$ is an integer

$\frac{x}{6} = n \pm \frac{1}{3}$

$x = 6 n \pm 2$

i.e. $x = \left\{\ldots \ldots \ldots - 16 , - 14 , - 10 , - 8 , - 4 , - 2 , 2 , 4 , 8 , 10 , 14 , 16 , \ldots \ldots \ldots\right\}$

graph{(tan(pix/6))^2-3 [-16, 16, -8, 8]}