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Sagot :
Answer:
Roots are -π/2 and π/2
Step-by-step explanation:
[tex]{ \bf{f(x) = 3 \cos(x) }}[/tex]
when x is -2π:
[tex]{ \sf{f( - 2\pi) = 3 \cos( - 2\pi) }} \\ { \sf{ = 3}}[/tex]
hence -2π is not a zero of the function
when x is 2π:
[tex]{ \sf{f(2\pi) = 3 \cos(2\pi) }} \\ { \sf{ = 3}}[/tex]
hence 2π is not a zero of the function
when x is π/2:
[tex]{ \sf{f( \frac{\pi}{2}) = 3 \cos( \frac{\pi}{2} ) }} \\ { \sf{ = 0}}[/tex]
Hence ±π/2 is the zero of the function.
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