Answer:
The zeros of the function in the interval of [-2π,2π] are
[tex]\pm \frac{\pi}{4},\pm \frac{3\pi}{4},\pm \frac{5\pi}{4},\pm \frac{7\pi}{4}[/tex].
Step-by-step explanation:
The given function is
[tex]f(x)=\frac{1}{2}\cos (2x)[/tex]
We have to find the zeros of the function in the interval of [-2π,2π].
If [tex]\cos x=0[/tex], then
[tex]x=\frac{(2n-1)\pi}{2}[/tex]
Where n is an integer.
Put f(x)=0, to find the zeroes of the function.
[tex]0=\frac{1}{2}\cos (2x)[/tex]
[tex]0=\cos (2x)[/tex]
[tex]2x=\pm \frac{\pi}{2},\pm \frac{3\pi}{2},\pm \frac{5\pi}{2},\pm \frac{7\pi}{2}[/tex] [tex][\because x\in [-2\pi,2\pi]\Rightarrow 2x\in [-4\pi,4\pi]][/tex]
Divide both sides by 2.
[tex]x=\pm \frac{\pi}{4},\pm \frac{3\pi}{4},\pm \frac{5\pi}{4},\pm \frac{7\pi}{4}[/tex]
Therefore the zeros of the function in the interval of [-2π,2π] are
[tex]\pm \frac{\pi}{4},\pm \frac{3\pi}{4},\pm \frac{5\pi}{4},\pm \frac{7\pi}{4}[/tex].
