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

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

## This Post Has 3 Comments

1. haileywilliams1 says:

Consider the function $f(x)=-4cos(x- \frac{ \pi }{2})$

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

so we solve

$-4cos(x- \frac{ \pi }{2})=0$

$cos(x- \frac{ \pi }{2})=0$

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)
$x- \frac{ \pi }{2}= \frac{ \pi }{2}$

$x= \pi$

and

ii)
$x- \frac{ \pi }{2}= 3\frac{ \pi }{2}$

$x=3\frac{ \pi }{2} +\frac{ \pi }{2}=2 \pi$

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

x=0, x=π, and x=2π

2. thegreentnt5025 says:

Recall that to get the x-intercepts, we set the f(x) = y = 0, thus

$\bf \stackrel{f(x)}{0}=-4cos\left(x-\frac{\pi }{2} \right)\implies 0=cos\left(x-\frac{\pi }{2} \right)&10;\\\\\\&10;cos^{-1}(0)=cos^{-1}\left[ cos\left(x-\frac{\pi }{2} \right) \right]\implies cos^{-1}(0)=x-\cfrac{\pi }{2}&10;\\\\\\&10;x-\cfrac{\pi }{2}=&10;\begin{cases}&10;\frac{\pi }{2}\\\\&10;\frac{3\pi }{2}&10;\end{cases}$

$\bf -------------------------------\\\\&10;x-\cfrac{\pi }{2}=\cfrac{\pi }{2}\implies x=\cfrac{\pi }{2}+\cfrac{\pi }{2}\implies x=\cfrac{2\pi }{2}\implies \boxed{x=\pi }\\\\&10;-------------------------------\\\\&#10;x-\cfrac{\pi }{2}=\cfrac{3\pi }{2}\implies x=\cfrac{3\pi }{2}+\cfrac{\pi }{2}\implies x=\cfrac{4\pi }{2}\implies \boxed{x=2\pi }$

3. yselahernandez02 says:

The function is $f(x)=-4cos( x-\frac{ \pi }{2} )$,

the x-intercepts of the function are the values of x, for which f(x)=0.

So we solve $-4cos( x-\frac{ \pi }{2} )=0$, where x∈[0,2π]

$-4cos( x-\frac{ \pi }{2} )=0$

$cos( x-\frac{ \pi }{2} )=0$

CosA=0, when A is $\frac{ \pi }{2} k$, where k is an integer.

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check the unit circle, the cosine, that is the first coordinate, is 0 at 90°, that is π/2, at 270°, that is π/2*3
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thus,

$x-\frac{ \pi }{2}$ is an element of $\frac{ \pi }{2} k$, that is

${...-2*\frac{ \pi }{2}, -1*\frac{ \pi }{2}, 0*\frac{ \pi }{2}, 1*\frac{ \pi }{2}, 2*\frac{ \pi }{2}, 3*\frac{ \pi }{2}...}$

that is

${...- \pi , -\frac{ \pi }{2}, 0, \frac{ \pi }{2}, \pi , \frac{ 3\pi }{2}, 2 \pi ...}$

this means that x is an element of:

${...- \pi+\frac{ \pi }{2} , -\frac{ \pi }{2}+\frac{ \pi }{2}, 0+\frac{ \pi }{2}, \frac{ \pi }{2}+\frac{ \pi }{2}, \pi+\frac{ \pi }{2} , \frac{ 3\pi }{2}+\frac{ \pi }{2}, 2 \pi+\frac{ \pi }{2} ...}$

so x is an element of

${...-\frac{ \pi }{2} , 0, \frac{ \pi }{2}, \pi , \frac{ 3\pi }{2} , 2 \pi,...}$

Since x can be from x=0 to x= 2π,

then the solution set is {0, π/2, π, 3π/2, 2π}

{0, π/2, π, 3π/2, 2π}

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