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

$\begin{array}{l}3+3\cos \mathrm{\theta }=2-2{\cos }^2\mathrm{\theta }\\ \Rightarrow 2{\cos }^2\mathrm{\theta }+3\cos \mathrm{\theta }+1=0\\ \Rightarrow (2\cos \mathrm{\theta }+1)(\cos \mathrm{\theta }+1)=0\end{array}$

$\therefore \cos \mathrm{\theta }=-1\text{ or }\cos \mathrm{\theta }=-\frac{1}{2}$

If $\cos \mathrm{\theta }=-1\text{ then }\mathrm{\theta }=\mathrm{\pi },3\mathrm{\pi },5\mathrm{\pi },7\mathrm{\pi },9\mathrm{\pi },\dots$

If $\cos \mathrm{\theta }=-1/2\text{ then }\mathrm{\theta }=\frac{2\mathrm{\pi }}{3},\frac{4\mathrm{\pi }}{3},\frac{8\mathrm{\pi }}{3},\frac{10\mathrm{\pi }}{3},\frac{14\mathrm{\pi }}{3},\dots$

Solutions in increasing order are 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2\mathrm{\pi }}{3}<\mathrm{\pi }<\frac{4\mathrm{\pi }}{3}<\frac{8\mathrm{\pi }}{3}<3\mathrm{\pi }<\frac{10\mathrm{\pi }}{3}<\frac{14\mathrm{\pi }}{3}<5\mathrm{\pi },\dots \\ \therefore {\mathrm{\theta }}_3+{\mathrm{\theta }}_7=\frac{4\mathrm{\pi }}{3}+\frac{14\mathrm{\pi }}{3}=6\mathrm{\pi }\end{array}$