Question

Where are the x-intercepts for f(x) = 4 cos(2x − π) from x = 0 to x = 2π?

x = pi over 4, x = 3 pi over 4, x = 5 pi over 4
x = pi over 2, x = 3 pi over 2
x = pi over 4, x = 3 pi over 4, x = 5 pi over 4, x = 7 pi over 4
x = pi over 2, x = 3 pi over 2, x = 5 pi over 2

Answer 1

@agent0smith

Answer 2

x-int means y=0, so you have to solve

4 cos(2x − π) = 0

Answer 3

x = 3pi/ 4

Answer 4

There's gonna be more than one.

Answer 5

Think about where cos theta = 0. You'll have to list several solutions, or use the +2*k*pi, where k is an integer.

Then set 2x − π equal to those, and solve for x.

Answer 6

is the one i said one of the options atleast ? so i know it could be A or C ..

Answer 7

Yeah the one you said is right

Answer 8

ok i dont know..

Answer 9

I'm going to take a guess and say A.

Answer 10

You could just plug in the other options and see if they work too...

or use the fact cos theta equals zero when theta is either
pi/2 + 2 k pi
or
3pi/2 + 2 k pi

then

2x − π = pi/2 +2 k pi
or
2x − π = 3pi/2 +2 k pi

and solve for x, and plug in values of k like 0, 1, 2 ,3... until you have all the ones from 0 to 2pi

Answer 11

Ohhhh... ok thank you.