Question

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

Answer 1

consider the function
`f(x)=-4cos(x- ( \pi )/(2))`

the x-intercepts of the graph of f, are the zeros of f, that is the solutions of f(x)=0.

so we solve


`-4cos(x- ( \pi )/(2))=0`


`cos(x- ( \pi )/(2))=0`

at this point we ask ourselves, keeping the unit circle in mind, "cosine of what is 0?"

In the unit circle, from 0 to 2π radians,

cos 90° = cos (π/2) rad = 0

and

cos 270° = cos (3π/2) rad = 0

this means that

i)

`x- ( \pi )/(2)= ( \pi )/(2)`


`x= \pi`

and

ii)

`x- ( \pi )/(2)= 3( \pi )/(2)`


`x=3( \pi )/(2) +( \pi )/(2)=2 \pi`

ii) also, check that
`f(0)=-4cos(0- ( \pi )/(2))=-4cos(-( \pi )/(2))` is also 0, because cos (-pi/2)= cos (-90°)=cos 270°=0


Answer:

x=0, x=π, and x=2π