Where are the x-intercepts for f(x) = 4 cos(2x − π) from x = 0 to x = 2π? a) x = pi over 4, x = 3 pi over 4, x = 5 pi over 4 b) x = pi over 2, x = 3 pi over 2 c) x = pi over 4, x = 3 pi over 4, x = 5 pi over 4, x = 7 pi over 4 d) x = pi over 2, x = 3 pi over 2, x = 5 pi over 2

The answer is choice C
The roots or x intercepts are:
x = pi/4
x = 3pi/4
x = 5pi/4
x = 7pi/4
There are four roots in the interval [0, 2pi]



Explanation:

We can rule out x = pi/2 as a solution because plugging in this x value leads to
f(x) = 4*cos(2*x-pi)
f(pi/2) = 4*cos(2*pi/2-pi)
f(pi/2) = 4*cos(pi-pi)
f(pi/2) = 4*cos(0)
f(pi/2) = 4*1
f(pi/2) = 4
The result isn't zero like we want, so x = pi/4 is not a root of f(x).
A root of f(x) would make f(x) = 0.

Since x = pi/2 is ruled out, this crosses choice B and choice D off the list.



The answer is between A and C. 

The two solution sets are nearly identical except for one x value: x = 7pi/4
Let's see if this x value leads to f(x) = 0 or not.
f(x) = 4*cos(2*x-pi)
f(7pi/4) = 4*cos(2*7pi/4-pi)
f(7pi/4) = 4*cos(14pi/4-pi)
f(7pi/4) = 4*cos(7pi/2-pi)
f(7pi/4) = 4*cos(7pi/2-2pi/2)
f(7pi/4) = 4*cos(5pi/2)
f(7pi/4) = 4*0
f(7pi/4) = 0

Since f(x) = 0 is true for this x value, this means x = 7pi/4 is a root of f(x)

So choice C is the most complete listing out all of the solutions in the interval [0,2pi]. 

That's why choice C is the answer

Note: [0,2pi] is interval notation indicating "start at 0, end at 2pi, include both endpoints"