Question

Find the zeros of the function in the interval of [-2pi,2pi]
f(x)=1/2cos2x

Answer 1

`\frac12\cos2x=0\implies\cos2x=0`

Since the cosine is zero when the angle is an odd multiple of
`\frac\pi2`, you have
`2x=\pm\frac\pi2,\pm\frac{3\pi}2,\pm\frac{5\pi}2,\pm\frac{7\pi}2`.

Divide both sides by 2 and you get
`x=\pm\frac\pi4,\pm\frac{3\pi}4,\pm\frac{5\pi}4,\pm\frac{7\pi}4`

Answer 2

The zeros of the function in the interval of [-2π,2π] are

`\pm (\pi)/(4),\pm (3\pi)/(4),\pm (5\pi)/(4),\pm (7\pi)/(4)`.

Explanation:

The given function is

`f(x)=(1)/(2)\cos (2x)`

We have to find the zeros of the function in the interval of [-2π,2π].

If
`\cos x=0`, then

`x=((2n-1)\pi)/(2)`

Where n is an integer.

Put f(x)=0, to find the zeroes of the function.

`0=(1)/(2)\cos (2x)`

`0=\cos (2x)`

`2x=\pm (\pi)/(2),\pm (3\pi)/(2),\pm (5\pi)/(2),\pm (7\pi)/(2)`
`[\because x\in [-2\pi,2\pi]\Rightarrow 2x\in [-4\pi,4\pi]]`

Divide both sides by 2.

`x=\pm (\pi)/(4),\pm (3\pi)/(4),\pm (5\pi)/(4),\pm (7\pi)/(4)`

Therefore the zeros of the function in the interval of [-2π,2π] are

`\pm (\pi)/(4),\pm (3\pi)/(4),\pm (5\pi)/(4),\pm (7\pi)/(4)`.