Let k be sum of all x in the interval $[0,2\mathrm{\pi }]$ such that $3{\cot }^2\mathrm{x}+8\cot \mathrm{x}+3=0$ then the value of $\frac{\mathrm{k}}{\mathrm{\pi }}$ is

$\begin{array}{r}3{\cot }^2\mathrm{x}+8\cot \mathrm{x}+3=0\\ \therefore \cot \mathrm{x}=\frac{-4\pm \sqrt{7}}{3}\end{array}$

Both roots are negative.

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{\pi }}{2}<{\mathrm{x}}_1,{\mathrm{x}}_2<\mathrm{\pi }\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\pi }<{\mathrm{x}}_1+{\mathrm{x}}_2<2\mathrm{\pi }\end{array}$

Product of roots $=\cot {\mathrm{x}}_1\cot {\mathrm{x}}_2=1$

$\begin{array}{l}\Rightarrow \cot {\mathrm{x}}_1\cot \left(\frac{3\mathrm{\pi }}{2}-{\mathrm{x}}_1\right)=1\\ \Rightarrow {\mathrm{x}}_1+{\mathrm{x}}_2=\frac{3\mathrm{\pi }}{2}\end{array}$

Similarly, ${\mathrm{x}}_1+{\mathrm{x}}_2=\frac{7\mathrm{\pi }}{2}$