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

Question

4476 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-18T07:53:21

$x={.........-16,-14,-10,-8,-4,-2,2,4,8,10,14,16,.........}$

$y = tan^2((pix)/6) -3$

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

$0 = tan^2((pix)/6) - 3$

or $3 = tan^2((pix)/6)$

$tan((pix)/6)=+-sqrt3$

$tan^-1(+-sqrt3) = (pix)/6$ and as $tan(+-pi/3)=+-sqrt3$

$(pix)/6=npi+-pi/3$ ,, where $n$ is an integer

$x/6=n+-1/3$

$x=6n+-2$

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

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

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