Let $0<{\mathrm{\theta }}_1<{\mathrm{\theta }}_2<{\mathrm{\theta }}_3<\dots$.... denote the positive solution of the eqution $3+3\cos \mathrm{\theta }=2{\sin }^2\mathrm{\theta }$The value of ${\mathrm{\theta }}_3+{\mathrm{\theta }}_7$is ______

$\begin{array}{l}2co{s}^2\theta +3cos\theta +1=0\\ \Rightarrow \therefore cos\theta =-1 or cos\theta =-\frac{1}{2}\end{array}$
Solution in increasing order are 
$\begin{array}{l}0<\frac{2\mathrm{\pi }}{3}<\mathrm{\pi }<\frac{4\mathrm{\pi }}{3}<\frac{8\mathrm{\pi }}{3}<3\mathrm{\pi }<\frac{10\mathrm{\pi }}{3}<\frac{14\mathrm{\pi }}{3}<\dots \dots \\ \Rightarrow \therefore {\mathrm{\theta }}_3+{\mathrm{\theta }}_7=\frac{4\mathrm{\pi }}{3}+\frac{14\mathrm{\pi }}{3}=6\mathrm{\pi }\end{array}$