Question

Find the zeros of the function in the interval of [-2pi,2pi]
f(x)=3cosx

Answer 1

`3\cos x=0\implies \cos x=0`

This happens when
`x` is an odd multiple of
`\frac\pi2`, and there are four instances of this happening in the given interval:
`\pm\frac\pi2,\pm\frac{3\pi}2`.

Answer 2

`x=-(3\pi)/(2),(\pi)/(2),(\pi)/(2),(3\pi)/(2)`

Explanation:

We have to find the zeros of the function f(x)=3cosx at [-2pi,2pi]

For zeros, we have f(x) = 0

`3\cos x=0`

Divide both sides by 3

`\cos x=0`

The value of cos is zero at
`(2n+1)(\pi)/(2)`

Hence, in the interval [-2pi,2pi], the value of x should be

`x=-(3\pi)/(2),(\pi)/(2),(\pi)/(2),(3\pi)/(2)`