# Solution of the inequality $\cos(2x)\geq 0$. And in general for $\cos(mx)$ with $m\in \Bbb R$?

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?

• Say you are solving $cos(6x)\ge0$. Then you'll need to draw 6 equi-distance and equi-length red arcs within a period of $2\pi$. Commented Mar 28, 2021 at 22:14

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!

• Excuse me for my delay....thank you very much and +1 and green check mark. Commented Apr 5, 2021 at 21:34