Question

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

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

x = pi over 2, x = 3 pi over 2

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

nobody has answered this yet i cant find the formula @phi do you know it?

Answer 2

the x-intercepts means what is the x value when f(x) value is zero.

in other words, find all the x values so that
4 cos(2x − π) = 0

Answer 3

and then when they =2 aswell?

Answer 4

you could simplify the function.
for example divide both sides by 4 to get
cos(2x − π) = 0/4
which is just
cos(2x − π) = 0

Answer 5

and that gets cos simplified

Answer 6

***and then when they =2 as well?***
I don't understand this question. Please ask a different way

Answer 7

ok.. u said i need to find when x=0.. i see that is an intercept in the question would i find it for when x is 2 because of the 2 intercept in the problem

Answer 8

to answer the question
cos(2x − π) = 0

you can either
1) solve for x (i.e. find the x's that work)
or
2) test the choices given as possible answers. Use a calculator and be sure to be in radian mode.

Answer 9

what exactly do you mean by radian mode?

Answer 10

to solve, we can use
cos(A-B)= cos(A)cos(B) + sin(A)sin(B)
and write
cos(2x-pi) = cos(2x)cos(pi) + sin(2x)sin(pi)

you should(?) know sin(pi) is 0, and cosI(pi) is -1
so you get
cos(2x-pi) = -cos(2x)
setting equal to 0
-cos(2x)= 0
or multiply both sides by -1 to get just
cos(2x)= 0

what angles is cos(y) zero? pi/2, pi/2 + pi, pi/2+2pi, etc
set 2x equal to each of these, and solve for x