if 0 $\leq x \leq 4$, and cos( $\pi\times\frac{x}{3})cos(\pi\times\frac{x}{2} ) \leq 0$ then find the range of x [closed]

I thought about trying each possibility but I think that it will take a lot of time if I do that. I tried to take to convert it to AM-GM like $\sqrt{\cos\left(\pi\times\frac{x}{3}\right)\cos\left(\pi\times\frac{x}{2}\right)} \leq AM$ but it just make the problem worse. thanks in advance. I think that maybe I can search for the possibility where one of them are negative and one of them are positive but i think that that is too hard

Since $0\le x \le 4$ there aren't many cases to check. Recall that $\cos(\theta)>0$ on $\left(2\pi n -\frac{\pi}{2},2\pi n +\frac{\pi}{2} \right)$ for $n \in \mathbb{N}$.

Using the previously mentioned, we see that $\cos\left( \frac{\pi}{3}x\right)$ is positive on $\left(-\frac{3}{2}, \frac{3}{2}\right)$ and negative on $\left(\frac{3}{2}, \frac{9}{2}\right)$ which covers all values of $0\le x\le 4$. Similarly for $\cos\left( \frac{\pi}{2}x\right)$ this one is positive on $(-1,1)$, negative on $(1,3)$, positive again on $(3,5)$.

Now, since on $\left(0, \frac{3}{2}\right)$ we know $\cos\left( \frac{\pi}{3}x\right)>0$ we just need to find what values in $\left(0, \frac{3}{2}\right)$ make $\cos\left( \frac{\pi}{2}x\right)<0$, but that is just $ (1,3) \cap \left(0, \frac{3}{2}\right)\cap (0,4) = \color{blue}{\left(1, \frac{3}{2}\right)}$. An analogous analysis on the interval $\left(\frac{3}{2}, \frac{9}{2}\right)$ tells you that $\left(\frac{3}{2}, \frac{9}{2}\right)\cap (3,5)\cap (0,4) = \color{blue}{\left(3, 4\right)}$ is also a solution set.

So the whole solution set is just the union of the sets we found before: $$ x \in \left(1, \frac{3}{2}\right) \cup \left(3,4\right) $$