If the largest possible value of  $x$ such that $0<x<\pi$  satisfying the equation  $\frac{2\sin 2x+1}{\cos x+\sin x}=\sqrt{2}\text{ is }\frac{a\pi }{b}.$ Find  $a+b$ $(Where a , b are co-prime )$

$$\begin{gathered} 2[{(\cos x+\sin x)}^2-1]+1=\sqrt{2}(\cos x+\sin x) \\ pur\cos x+\sin x=t \\ 2t^2-1=\sqrt{2}t\Rightarrow 2t^2-\sqrt{2}t-1=0 \\ t=\frac{\sqrt{2}\pm \sqrt{10}}{4}=\frac{\sqrt{2}(1\pm \sqrt{5})}{4} \\ \cos x+\sin x=\frac{\sqrt{2}(1-\sqrt{5})}{4} \\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.(\cos x+\sin x)=-\frac{(\sqrt{5}-1)}{4} \\ \sin (x+\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.)=\sin \left(\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)\text{ or }\sin \raisebox{1ex}{$11\pi $}\!\left/ \!\raisebox{-1ex}{$10$}\right. \\ x+\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.=\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$10$}\right.\text{or }\raisebox{1ex}{$11\pi $}\!\left/ \!\raisebox{-1ex}{$10$}\right. \\ x=\raisebox{1ex}{$11\pi $}\!\left/ \!\raisebox{-1ex}{$10$}\right.-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.=\frac{17\pi }{20} \\ a=17,b=20 \\ a+b=37 \end{gathered}$$