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The given function is a cosine function that has an output of 0 when the the input is [tex]\displaystyle \frac{\pi}{2}[/tex].
Reasons:
The given function is f(x) = 5·cos(x)
The characteristics of the function are;
The operator of the function is the cosine function
The general form of the cosine function is; y = A·cos(ω·x - Φ) + k
Where:
A = The amplitude of the function
[tex]\displaystyle The \ period = \mathbf{ \frac{2 \cdot \pi}{\omega}}[/tex]
[tex]\displaystyle The \ phase \ shift = \frac{\phi}{\omega}[/tex]
The vertical shift = k
Therefore, by comparison, we have;
The amplitude, A = 5
The period of the function = 2·π
The phase shift, Ф = 0
The vertical shift, k = 0
The zero of the function are given when the output of the function is 0,
which is found as follows;
[tex]f(x) = \mathbf{5 \cdot cos(x) }= 0[/tex]
cos(x) = 0
[tex]\displaystyle x = arcos(0) = \frac{\pi }{2}[/tex]
Which gives a zero of the function as; [tex]\displaystyle \underline{\left(\frac{\pi}{2} , \, 0 \right)}[/tex]
Learn more about the cosine function here:
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