How do you find the x intercepts of $y=sec^4((pix)/8)-4$ ?

Question

3080 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-31T13:25:49

$x=4n-2$ , $n=1,2,3,$ ...

$y=sec^4((pix)/8)-4$

at $x$ intercepts, $y=0$
$sec^4((pix)/8)-4=0$

$(sec^2((pix)/8)-2)(sec^2((pi x)/8)+2)=0$

$(sec^2((pi x)/8)+2)=0$ , $sec^2((pi x)/8)=-2$ , cannot solve.

$sec^2((pix)/8)-2=0$ , $sec((pix)/8)=+-sqrt2$

$(pix)/8=pi/4, (3pi)/4, (5pi)/4,...$

$x=2,6,10,...$
$x=4n-2$ , $n=1,2,3,$ ...

How do you find the x intercepts of y=sec^4((pix)/8)-4y=sec^4((pix)/8)-4?