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1. 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

    

   x=0, x=π, and x=2π

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

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

   [tex]\bf -------------------------------\\\\
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 }\\\\
-------------------------------\\\\
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 }[/tex]

   Reply

3. The function is [tex]f(x)=-4cos( x-\frac{ \pi }{2} )[/tex], 

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

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

   [tex]-4cos( x-\frac{ \pi }{2} )=0[/tex]

   [tex]cos( x-\frac{ \pi }{2} )=0[/tex]

   CosA=0, when A is [tex]\frac{ \pi }{2} k[/tex], where k is an integer.

   -------------------------------------------------------------------------------------------------
   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
   -------------------------------------------------------------------------------------------------

   thus, 

   [tex]x-\frac{ \pi }{2}[/tex] is an element of [tex]\frac{ \pi }{2} k[/tex], that is

   [tex]{...-2*\frac{ \pi }{2}, -1*\frac{ \pi }{2}, 0*\frac{ \pi }{2}, 1*\frac{ \pi }{2}, 2*\frac{ \pi }{2}, 3*\frac{ \pi }{2}...}[/tex]

   that is

   [tex]{...- \pi , -\frac{ \pi }{2}, 0, \frac{ \pi }{2}, \pi , \frac{ 3\pi }{2}, 2 \pi ...}[/tex]

   this means that x is an element of:

   [tex]{...- \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} ...}[/tex]

   so x is an element of

   [tex]{...-\frac{ \pi }{2} , 0, \frac{ \pi }{2}, \pi , \frac{ 3\pi }{2} , 2 \pi,...}[/tex]

   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}

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

   Reply