Question

Find the zeros of the function f(x)=1/2cos 2x with interval -2pi,2pi]

Answer 1

$\bf f(x)=\cfrac{1}{2}cos(2x)\qquad [-2\pi \ ,\ 2\pi ]\iff [0\ , \ 2\pi ]\\\\ -----------------------------\\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ \boxed{2cos^2(\theta)-1} \end{cases}\\\\ -----------------------------$

$\bf 0=\cfrac{cos(2x)}{2}\implies 0=cos(2x)\implies 0=2cos^2(x)-1 \\\\\\ \cfrac{1}{2}=cos^2(x)\implies \pm\sqrt{\cfrac{1}{2}}=cos(x)\implies \cfrac{1}{\pm\sqrt{2}}=cos(x) \\\\\\ \pm \cfrac{\sqrt{2}}{2}=cos(x)\implies \measuredangle x= \begin{cases} \frac{\pi }{4}\\\\ \frac{3\pi }{4}\\\\ \frac{5\pi }{4}\\\\ \frac{7\pi }{4} \end{cases}$

Answer 2

The period for Cos is 2pi/k.
In this function, k is 2, so the period is pi.
This means there will be zeros at pi/4, 3pi/4, 5pi/4, 7pi/4 on the position side and the same on the negative side. -pi/4, -3pi/4, -5pi/4, -7pi/4