Is there always a positive $x$ that satisfies $\cos(n_1x)\leq0$, $\cos(n_2x)\leq0$, $\cos(n_3x)\leq0$ for given distinct positive integers $n_i$? Prove or disprove:

Given distinct $n_1$, $n_2$, $n_3$ $\in \mathbb{N}$, \begin{cases} \cos(n_1x)\leq0 \\ \cos(n_2x)\leq0 \\ \cos(n_3x)\leq0\end{cases} has a
positive solution.

My first attempt was by writing each cosine in terms of $\cos(x)$, then by setting $\cos(x) = t$, it's solved like a normal system of inequalities.
This didn't work for me because writing each $\cos(n_jx)$ in terms of $\cos(x)$ gets harder as the $n_j$ increases, and I'm also unable to find a pattern behind their conversion, this doesn't let me find a generalized proof.
Second attempt was by using parametric functions:
$\\ \sin{(\alpha)}=\frac{2t}{1+t^2}\ \ \ \mbox{where }\ t=\tan{\left(\frac{\alpha}{2}\right)}\mbox{ and }\alpha\neq\pi+2k\pi$
$\\ \cos{(\alpha)}=\frac{1-t^2}{1+t^2}\ \ \ \mbox{where }\ t=\tan{\left(\frac{\alpha}{2}\right)}\mbox{ and }\alpha\neq\pi+2k\pi$
$\\ \tan{(\alpha)}=\frac{2t}{1-t^2}\ \ \ \mbox{where }\ t=\tan\left(\frac{\alpha}{2}\right)\mbox{ and }\alpha\neq\frac{\pi}{2}+k\pi\ \wedge\ \alpha\neq\pi+2k\pi$
Same problem as the first attempt, I'm unable to generalize their conversion in terms of a fixed $α$, so can't again find a generalized proof.
A third attempt was using proof by contradiction:
We suppose no such $x$ exists. Then, for every positive $x,$ at least one of the three cosines is always positive.
... Can't get past this point.
 A: 
For each set of positive integers $n_1,n_2,n_3,$
$$\cos(n_1x),\cos(n_2x),\cos(n_3 x)\leq0$$ has a real solution.

Observation: if the statement specifies greater than three positive integers instead, then it is false: a counterexample is $n_4=4.$
Proof (but missing Case 2)
It suffices to show that for ascendingly-ordered positive integers $a,b,c,$ $$\cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right),\cos\left(\frac\pi2cx\right)\leq0$$ has a real solution.
Noting that $$\cos\left(\frac\pi2nx\right)\leq0\\\iff\\ x\in\ldots\cup\left[-\frac3n,-\frac1n\right]\cup\left[\frac1n,\frac3n\right]\cup\left[\frac5n,\frac7n\right]\cup\left[\frac9n,\frac{11}n\right]\cup\ldots,$$ there are four possible cases:

*

*$c\leq 3a:$
\begin{align}&a<b &\text{and} &&b<c &&\text{and} &&a<c\leq3a<3c\\
   &\frac1b<\frac1a &\text{and} &&\frac3c<\frac3b &&\text{and}
   &&\frac1c<\frac1a\leq\frac3c<\frac3a \end{align}
$$[\frac1a,\frac3c]\subseteq[\frac1a,\frac3a]\cap[\frac1b,\frac3b]
   \cap[\frac1c,\frac3c]
   \\\cos\left(\frac\pi2ax\right), \cos\left(\frac\pi2bx\right),
   \cos\left(\frac\pi2cx\right)\leq0
   \quad\text{ on } \left[\frac1a,\frac3c\right].$$


*$\displaystyle b<3a<c<\frac{2ab}{3a-b}:$
This section needs to be filled in; for convenience: Desmos links 1 & 2. I have duplicated the next four lines from Case 3 only so that this Answer is at least a proof of the given statement were it to require only $\mathbf{\mathit{\cos(n_1x),\cos(n_2x)}}$ to be nonpositive.
$$a<b<3a<3b \\\frac1b<\frac1a<\frac3b<\frac3a \\
   [\frac1a,\frac3b]\subseteq[\frac1a,\frac3a]\cap[\frac1b,\frac3b]
   \\\cos\left(\frac\pi2ax\right), \cos\left(\frac\pi2bx\right)\leq0
   \quad\text{ on } \left[\frac1a,\frac3b\right].$$


*$b<3a<c$ and $\displaystyle c\geq\frac{2ab}{3a-b}:$
$$a<b<3a<3b \\\frac1b<\frac1a<\frac3b<\frac3a \\
   [\frac1a,\frac3b]\subseteq[\frac1a,\frac3a]\cap[\frac1b,\frac3b]
   \\\cos\left(\frac\pi2ax\right), \cos\left(\frac\pi2bx\right)\leq0
   \quad\text{ on } \left[\frac1a,\frac3b\right].$$
Since $\displaystyle \frac3b-\frac1a=\frac{3a-b}{ab}\geq\frac2c,$ which is the size of the smallest interval on which $\displaystyle\cos\left(\frac\pi2cx\right)$ must be somewhere nonpositive, some point in $\displaystyle\left[\frac1a,\frac3b\right]$ in fact contains a solution for $\cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right),\cos\left(\frac\pi2cx\right)\leq0.$


*$b\in\big[(2m{+}1)a,(2m{+}3)a\big]$ and $m\in\mathbb Z^+:$
\begin{align}b&\leq(2m+3)a &\text{and} &&b&\geq(2m+1)a\\
   \frac1a&\leq\frac{2m+3}b &\text{and}
   &&\frac3a&\geq\frac{6m+3}b\\&\leq\frac{4m+1}b &&&&>\frac{4m+3}b\\
   &\frac1a\leq\frac{4m+1}b<\frac{4m+3}b<\frac3a\end{align}
$$\left[\frac{4m+1}b,\frac{4m+3}b\right]\subseteq
   \left[\frac1a,\frac3a\right]\cap\left[\frac{4m+1}b,\frac{4m+3}b\right]\\
   \cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right)≤0
   \quad\text{ on } \left[\frac{4m+1}b,\frac{4m+3}b\right].$$
Since $\displaystyle\frac{4m+3}b-\frac{4m+1}b=\frac2b>\frac2c,$ which is the size of the smallest interval on which $\displaystyle\cos\left(\frac\pi2cx\right)$ must be somewhere nonpositive, some point in $\displaystyle\left[\frac{4m+1}b,\frac{4m+3}b\right]$ in fact contains a solution for $\cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right),\cos\left(\frac\pi2cx\right)\leq0.$
A: Let's assume $n_1<n_2<n_3$. We know that for $\frac{π}{2n_1}≤x≤\frac{3π}{2n_1}$, $\cos(n_1x)≤0$.
Now consider $\cos(n_2x)$. $\frac{π}{2n_2}<\frac{π}{2n_1}$ as $n_1<n_2$. If $\frac{3π}{2n_2}>\frac{π}{2n_1}$ we would immediately have a region where both $\cos(n_1x)≤0$ and $\cos(n_2x)≤0$, viz. $\frac{π}{2n_1}≤x≤\frac{3π}{2n_2}$. Thus let's assume the opposite, that $\frac{3π}{2n_2}<\frac{π}{2n_1}$ $\implies 3n_1<n_2$.
Now, the next interval on which $\cos(n_2x)≤0$ is $\frac{5π}{2n_2}≤x≤\frac{7π}{2n_2}$. But, $n_2>3n_1>\frac{5n_1}{3}$ $\implies 5n_1<3n_2 \implies \frac{5π}{2n_2}<\frac{3π}{2n_1}$. Thus for $\frac{5π}{2n_2}≤x≤\frac{3π}{2n_1}$ both $\cos(n_1x)≤0$ and $\cos(n_2x)≤0$.
Similarly we can argue for $\frac{5π}{2n_1}<x<\frac{7π}{2n_1}$. This may seem redundant, but notice that we can repeat our entire argument but this time take $n_2$ and $n_3$ instead. The interval $\frac{π}{2n_1}<x<\frac{3π}{2n_1}$ contains $\frac{5π}{2n_2}<x<\frac{7π}{2n_2}$. Now in this interval apply this argument but for $n_2$ and $n_3$ instead.

Similarly we can argue for $\frac{5π}{2n_1}<x<\frac{7π}{2n_1}$

From this we can also see why the claim would not be true for $4$ $n$s (Hint: for the fourth $n$ we will have to look at the next interval $\frac{9π}{2n_1}<x<\frac{11}{2n_1}$, for which the ratio $\frac{5}{3}$ we got will change to something bigger than $3$)
