SOLUTION: Where are the x-intercepts for f(x) = 4 cos(2x − π) from x = 0 to x = 2π? I have no idea where to begin. On WolframAlpha, it says the x int's are npi/2 - pi/4, w

Algebra ->  Trigonometry-basics -> SOLUTION: Where are the x-intercepts for f(x) = 4 cos(2x − π) from x = 0 to x = 2π? I have no idea where to begin. On WolframAlpha, it says the x int's are npi/2 - pi/4, w      Log On


   

Question 876351: Where are the x-intercepts for f(x) = 4 cos(2x − π) from x = 0 to x = 2π?
I have no idea where to begin. On WolframAlpha, it says the x int's are npi/2 - pi/4, where n is an element of the integers. I do not know how they arrived at that answer.
I know the amplitude is 4, the period is pi, but when I solve bx - h = 0 I get pi/2, but not pi/2 - pi/4. How do they obtain this answer?
Many thanks for your help
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
cos%28z%29=0 when z=%281%2F2%29pi and z=%283%2F2%29pi
So then,
2x-pi=%281%2F2%29pi and 2x-pi=%283%2F2%29pi
2x=%283%2F2%29pi and 2x=%285%2F2%29pi
x=%283%2F4%29pi and x=%285%2F4%29pi
And since cosine is periodic in 2pi
x=%283%2F4%29pi+%2B-+2pi%2An and x=%285%2F4%29pi+%2B-+2pi%2An
You can combine those two together to get the Wolfram answer.