Sum of the solutions in  $[0,8\pi ]$ of the equation $\tan x+\cot x+1=\cos \left(x+\frac{\pi }{4}\right)$ is $k\pi$  then 2K = 

$\tan x+\cot x=\cos \left(x+\frac{\pi }{4}\right)-1$

$LHS\le -2$  or  $\ge 2$  

$\therefore$$RHS\le -2$    
$\therefore LHS=RHS=-2$

$\therefore \cos \left(x+\frac{\pi }{4}\right)-1=-2\Rightarrow \cos \left(x+\frac{\pi }{4}\right)=-1$

$$\begin{gathered} \Rightarrow x+\frac{\pi }{4}=2n\pi +\pi \\ \Rightarrow x=2n\pi +\frac{3\pi }{4} \end{gathered}$$

$\Rightarrow x=\frac{3\pi }{4},\frac{11\pi }{4},\frac{19\pi }{4},\frac{27\pi }{4}$

Sum =  $15\pi$