Respuesta :
consider the function [tex]f(x)=-4cos(x- \frac{ \pi }{2})[/tex]
the x-intercepts of the graph of f, are the zeros of f, that is the solutions of f(x)=0.
so we solve
[tex]-4cos(x- \frac{ \pi }{2})=0[/tex]
[tex]cos(x- \frac{ \pi }{2})=0[/tex]
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)
[tex]x- \frac{ \pi }{2}= \frac{ \pi }{2}[/tex]
[tex]x= \pi [/tex]
and
ii)
[tex]x- \frac{ \pi }{2}= 3\frac{ \pi }{2} [/tex]
[tex]x=3\frac{ \pi }{2} +\frac{ \pi }{2}=2 \pi [/tex]
ii) also, check that [tex]f(0)=-4cos(0- \frac{ \pi }{2})=-4cos(-\frac{ \pi }{2})[/tex] is also 0, because cos (-pi/2)= cos (-90°)=cos 270°=0
Answer:
x=0, x=π, and x=2π
the x-intercepts of the graph of f, are the zeros of f, that is the solutions of f(x)=0.
so we solve
[tex]-4cos(x- \frac{ \pi }{2})=0[/tex]
[tex]cos(x- \frac{ \pi }{2})=0[/tex]
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)
[tex]x- \frac{ \pi }{2}= \frac{ \pi }{2}[/tex]
[tex]x= \pi [/tex]
and
ii)
[tex]x- \frac{ \pi }{2}= 3\frac{ \pi }{2} [/tex]
[tex]x=3\frac{ \pi }{2} +\frac{ \pi }{2}=2 \pi [/tex]
ii) also, check that [tex]f(0)=-4cos(0- \frac{ \pi }{2})=-4cos(-\frac{ \pi }{2})[/tex] is also 0, because cos (-pi/2)= cos (-90°)=cos 270°=0
Answer:
x=0, x=π, and x=2π
