$\newcommand\al\alpha$This answer is incomplete -- it reduces the problem to a conjecture involving, not $x$, but only 
$$u:=\sum_{1\le i<j\le6}c_{ij}^2\quad\text{and}\quad 
v:=\sum_{1\le i<j<k\le6}c_{ij}c_{ik}c_{jk},$$
where $c_{ij}:=\cos\al_{ij}$. This answer is somewhat similar to the [previous answer][1] -- to a much simpler question (the previous answer was complete). 

Let $p(x)$ be the polynomial in question. Note that $0\le u\le\binom62=15$ and let 
$$x_*:=1-\sqrt{u/15}.$$
According to [this answer by Fedor Petrov][2], 
$$v\le\frac4{3\sqrt{15}}\,u^{3/2}.$$
So, 
$$p(x_*)=\frac4{3\sqrt{15}}\,u^{3/2}-v\ge0.$$
Also, $p(-\infty+)=-\infty<0$. 
So, $p$ has a root $\le1/2$ in the case when $x_*\le1/2$. 

It remains to consider the case when $x_*>1/2$, that is, when 
$0\le u<15/4$. Note that 
$$p(1/2)=-5/4+u-v.$$
According to [this conjecture][3], $u-v\ge23/8$ if $u\le15/4$. 
So, in the case when $x_*>1/2$ we have 
$$p(1/2)\ge-5/4+23/8=13/8\ge0.$$
So, provided the mentioned conjecture is true, $p$ will have a root $\le1/2$ in the case when $x_*>1/2$ as well. 

---

We also see that it would be enough to prove that $u-v\ge10/8$ (given $u\le15/4$), which seems a much weaker claim than $u-v\ge23/8$ (given $u\le15/4$). 

  [1]: https://mathoverflow.net/a/478921/36721
  [2]: https://mathoverflow.net/a/479088/36721
  [3]: https://mathoverflow.net/q/479107/36721