$1+\tan x=2\sqrt{2}\sin \left( x+\frac{\pi }{4} \right)\Rightarrow \frac{\operatorname{sinx}+\operatorname{cosx}}{\operatorname{cosx}}=2\sqrt{2}\left( \operatorname{sinx}\cos \frac{\pi }{4}+\operatorname{cosx}\sin \frac{\pi }{4} \right)$  $\operatorname{sinx}+\operatorname{cosx}=2\sqrt{2}\operatorname{cosx}\left( \frac{\sqrt{2}}{2}\operatorname{sinx}+\frac{\sqrt{2}}{2}\operatorname{cosx} \right)\Rightarrow \operatorname{sinx}+\operatorname{cosx}=2\operatorname{cosx}\left( \operatorname{sinx}+\operatorname{cosx} \right)$  $\Rightarrow \left( \operatorname{sinx}+\operatorname{cosx} \right)\left( 2\operatorname{cosx}-1 \right)=0$  $\xrightarrow{\left( k\in z \right)}\left\{ \begin{matrix}    \operatorname{sinx}+\operatorname{cosx}=0\Rightarrow tanx=-1\Rightarrow x=k\pi -\frac{\pi }{4}\Rightarrow {{x}_{1}}=\frac{3\pi }{4}  \\    2\operatorname{cosx}-1=0\Rightarrow \operatorname{cosx}=\frac{1}{2}\Rightarrow x=2k\pi \pm \frac{\pi }{3}\Rightarrow {{x}_{2}}=\frac{\pi }{3}  \\ \end{matrix}\Rightarrow {{x}_{1}}+{{x}_{2}}=\frac{3\pi }{4} \right.+\frac{\pi }{3}=\frac{13\pi }{3}$