`y = sec^4 ((pix)/8) - 4` Find the x-intercepts of the graph.

Textbook Question

Chapter 5, 5.3 - Problem 48 - Precalculus (3rd Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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x-intercepts are the points where y=0. So we have to solve equation

`sec^4((pix)/8)-4=0,` or `sec^4((pix)/8)=4.`

Then `sec^2((pix)/8)=2,` or `sec((pix)/8)=+-sqrt(2).`

sec(y) = 1/cos(y), therefore `cos((pix)/8)=+-1/sqrt(2).`

The general solution is

`(pix)/8 = +-arccos(1/sqrt(2))+2kpi` and `(pix)/8 = +-arccos(-1/sqrt(2))+2kpi,` `kinZZ.`

`arccos(1/sqrt(2))=pi/4,` `arccos(-1/sqrt(2)) = (3pi)/4.`

So the answer is

`x=+-2+16k,` `x=+-6+16k,` `kinZZ,` which can be written as

`x = 2+4k, kinZZ.`

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