Sum of all the solutions in $[0,4\pi ]$ of the equation $\tan x+\cot x+1=\cos \left(x+\frac{\pi }{4}\right)$ is

$\begin{array}{r}\tan \mathrm{x}+\cot \mathrm{x}+1=\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\\ \Rightarrow \tan \mathrm{x}+\cot \mathrm{x}=\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)-1\end{array}$
$\text{ Now }\tan x+\cot x\le -2\text{ and }\cos \left(x+\frac{\pi }{4}\right)-1\ge -2\text{. }$
It implies that equality holds when both are -2
$\begin{array}{l}\Rightarrow \cos \left(x+\frac{\pi }{4}\right)=-1\\ \Rightarrow x+\frac{\pi }{4}=(2m+1)\pi ,m\in Z\\ \Rightarrow x=\frac{3\pi }{4}\text{ or }\frac{11\pi }{4}\end{array}$
$\text{ Therefore, the sum of the solutions is }\frac{3\pi }{4}+\frac{11\pi }{4}=\frac{7\pi }{2}\text{. }$