If the equation $x^2+5+4\cos \left(\alpha +\beta \right)=2x$ has at least one solution where $\alpha ,\beta \in \left[2,5\right]$ then the value of $\alpha +\beta$ is equal to

$x^2+5+4\cos \left(\alpha +\beta \right)=2x$

$\Rightarrow {\left(x-1\right)}^2+4=-4\cos \left(ax+\beta \right)$

Here,  ${\left(x-1\right)}^2+4\ge 4,-4\cos \left(\alpha x+\beta \right)\le 4$$\left(\because -1\le \cos \left(\alpha x+\beta \right)\le 1\right)$

But, the given equation has at least one solution, then ${\left(x-1\right)}^2+4=4=-4\cos \left(\alpha x+\beta \right)$

$\therefore x=1\Rightarrow \cos \left(\alpha +\beta \right)=-1$

$\Rightarrow \alpha +\beta =3\pi \left(\because \alpha +\beta \in \left[4,10\right]\right)$