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