$\sin \mathrm{\theta }+\sqrt{3}\cos \mathrm{\theta }=6\mathrm{x}-{\mathrm{x}}^2-11,0\le \mathrm{\theta }\le 4\mathrm{\pi },\mathrm{x}\in \mathrm{R}$ holds for

$\begin{array}{r}\sin \mathrm{\theta }+\sqrt{3}\cos \mathrm{\theta }=-2-\left({\mathrm{x}}^2-6\mathrm{x}+9\right)=-2-(\mathrm{x}-3{)}^2\\ \therefore \sin \mathrm{\theta }+\sqrt{3}\cos \mathrm{\theta }\ge -2\text{ and }-2-(\mathrm{x}-3{)}^2\le -2\end{array}$
As a result, we have $\sin \mathrm{\theta }+\sqrt{3}\cos \mathrm{\theta }=-2$ and then x = 3
$\therefore \mathrm{x}=3\text{ and }\cos \left(\mathrm{\theta }-\frac{\mathrm{\pi }}{6}\right)=-1,\text{ i.e., }\mathrm{\theta }-\frac{\mathrm{\pi }}{6}=\mathrm{\pi },3\mathrm{\pi }$