Just I have a little doubt but that I think to have solved. If I consider that inequality
$$\cos(2x)\geq 0 \tag 1$$
if I put $2x=u$, we have
$$\cos(u)\geq 0 \iff -\frac{\pi}2+2\Bbb Z\pi \leq u \le \frac{\pi}2+2\Bbb Z\pi$$
that for the $(1)$ we will have:
$$-\frac{\pi}4+\Bbb Z\pi \leq x \le \frac{\pi}4+\Bbb Z\pi \tag 2$$
If I drew the arc only in the interval $-\frac{\pi}4\leq x \le \frac{\pi}4$ I would have made a very serious mistake because in the $(2)$ I have also a main periodicity of $\pi$ that I must to draw necessarily. If the main periodicity was different (it is not in this exercise-I'm supposing), for example $\pi/3$, I should always draw in the interval $[0,2\pi[$, the interval $(a,b)\subseteq[0,2\pi[$ without the periodicity ($\iff$ the continuous red arc), but also the other continuous red arc $(a+\frac{\pi}3,b+\frac{\pi}3)\subseteq[0,2\pi[$.
Reading the comment of the user @Hyenazixiao if I have $\cos(6x)≥0$ I'll need to draw $6$ equi-distance and equi-length red arcs within of a period of $2π$. Is there a proof for this event or can I see from the last solution looking the periodicity?
As an algebraic answer, note that in one period $2\pi$, $\cos(x)\geq0$ on $\left[0,\;\frac{\pi}{2}\right]\cup\left[\frac{3\pi}{2},\;2\pi\right)$. By multiplying the inside of the cosine by $m\in\mathbb{R}$, we are effectively compressing or "squishing" the function by a factor of $m$.
So, assuming $m\neq0$ (if $m$ were $0$ the inequality would hold everywhere) the new period is $\frac{2\pi}{m}$, and the new interval where $\cos(mx)\geq0$ in one period is now $\left[0,\;\frac{\pi}{2m}\right]\cup\left[\frac{3\pi}{2m},\;\frac{2\pi}{m}\right)$.
Finally, to account for periodicity, we add $\frac{2k\pi}{m}$ to each end of the two intervals, where $k\in\mathbb{Z}$. Hence, the set of $x$ satisfying the inequality $\cos(mx)\geq 0$ is
$$x\in\left[\frac{2k\pi}{m},\;\frac{\pi}{2m}+\frac{2k\pi}{m}\right]\cup\left[\frac{3\pi}{2m}+\frac{2k\pi}{m},\;\frac{2\pi}{m}+\frac{2k\pi}{m}\right)$$
which can be simplified to:
$$x\in\left[\frac{2k\pi}{m},\;\frac{(4k+1)\pi}{2m}\right]\cup\left[\frac{(4k+3)\pi}{2m},\;\frac{(2k+2)\pi}{m}\right).$$
Geometrically, for your red arcs, $\cos(mx)$ would have $m$ red arcs within the circle. Hope this helps!