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$\begin{aligned} \frac {1+2\sin 2x}{\sin x + \cos x} & = \sqrt 2 \\ \frac {1+2\cos \left(\frac \pi 2 - 2x\right)}{\sqrt 2 \cos \left(\frac \pi 4 - x\right)} & = \sqrt 2 \\ 1+2\cos \left(2\left(\frac \pi 4 - x\right) \right) & = 2 \cos \left(\frac \pi 4 - x\right) & \small \color{#3D99F6} \text{Let }\theta = \frac \pi 4 - x \\ 1 + 2\cos 2\theta & = 2\cos \theta \\ \cos \theta - \cos 2\theta & = \frac 12 \end{aligned}$

Note that $\displaystyle \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12$ (see proof ). A particular case is $2n+1=5$ , then $\cos \dfrac \pi 5 + \cos \dfrac {3\pi}5 = \dfrac 12$ .

We note that $\theta = \dfrac \pi 5$ is a solution, because then $\cos 2\theta = \cos \dfrac {2\pi} 5 = - \cos \dfrac {3\pi}5$ . $\implies \cos \theta - \cos 2\theta = \cos \dfrac \pi 5 + \cos \dfrac {3\pi}5 = \dfrac 12$ . And from $\theta = \dfrac \pi4 - x$ , $\implies x = \dfrac \pi{20}$ . Since $\cos \theta = \cos (- \theta)$ , then $x - \dfrac \pi 4 = \dfrac \pi 5 = \dfrac {9\pi}{20}$

We note that $\theta = \dfrac {3\pi}5$ is also a solution, because then $\cos 2\theta = \cos \dfrac {6\pi} 5 = - \cos \dfrac \pi 5$ . $\implies \cos \theta - \cos 2\theta = \cos \dfrac {3\pi}5 + \cos \dfrac \pi 5 = \dfrac 12$ . And from $\theta = \dfrac \pi4 - x$ , $\implies x = -\dfrac {7\pi}{20}$ . And $x - \dfrac \pi 4 = \dfrac {3\pi}5 = \dfrac {17\pi}{20}$ .

The largest $x < \pi$ is $x = \dfrac {17\pi}{20}$ , $\implies a+b = 17+20 = \boxed{37}$

$\frac{1+2\sin {2x}}{\sin{x}+\cos{x}}=\sqrt{2}$

$1+2\sin {2x}=\sqrt{2}(\sin{x}+\cos{x})$

$\frac{1}{2}(1+2\sin {2x})=\frac{\sqrt{2}}{2}(\sin{x}+\cos{x})$

$\frac{1}{2}+\sin {2x}=\frac{\sqrt{2}}{2}\sin{x}+\frac{\sqrt{2}}{2}\cos{x}$

$\frac{1}{2}+\cos {(\frac{\pi}{2}-2x)}=\sin{\frac{\pi}{4}}\sin{x}+\cos{\frac{\pi}{4}}\cos{x}$

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

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

Now we let $\theta=\frac{\pi}{4}-x$ .

$\frac{1}{2}+\cos{(2\theta)}=\cos{\theta}$

It then remains to make use of the hint given in the question:

$2\sin{\theta}(\frac{1}{2}+\cos{(2\theta)})=2\sin{\theta}\cos{\theta}$

$\sin{\theta}+2\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}$

$\sin{(2-1)\theta}+2\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}$

$\sin{(2\theta)}\cos{\theta}-\sin{\theta}\cos{(2\theta)}+2\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}$

$\sin{(2\theta)}\cos{\theta}+\sin{\theta}\cos{(2\theta)}=\sin{(2\theta)}$

$\sin{(2+1)\theta}=\sin{(2\theta)}$

$\sin{(3\theta)}=\sin{(2\theta)}$

$\sin{(3\theta)}=\sin{(\pi-2\theta)}$

$3\theta=\pi-2\theta$

$(3+2)\theta=\pi$

$5\theta=\pi$

$\theta=\frac{\pi}{5}$

Note that the above calculations was just to work out the basic solution for $\theta$ . Its full set of solutions is $\theta=\frac{\pi}{5}, -\frac{\pi}{5}, -\frac{3\pi}{5}$ and so on. This is because the sine function has a period of $2\pi$ so upon divison by five to get $\theta$ , the period becomes $\frac{2\pi}{5}$ . Note that $\theta=-\frac{3\pi}{5}$ is the value that gives the largest $x<\pi$ :

$\frac{\pi}{4}-x=-\frac{3\pi}{5}$

$x=\frac{3\pi}{5}+\frac{\pi}{4}$

$x=\frac{(12+5)\pi}{20}$

$x=\frac{17\pi}{20}$

Therefore $a=17, b=20$ and $a+b=\boxed{37}$

Just...

$\dfrac{1+2\sin 2x}{\sin x+\cos x} \\ =\dfrac{1+4\sin x\cos x}{\sin x+\cos x} \\ =\dfrac{2(1+2\sin x\cos x)}{\sin x+\cos x}-\dfrac{1}{\sin x+\cos x} \\ =\dfrac{2(\sin x+\cos x)^2}{\sin x+\cos x}-\dfrac{1}{\sin x+\cos x} \\ =2(\sin x+\cos x)-\dfrac{1}{\sin x+\cos x}=\sqrt{2}.$

Let $\sin x+\cos x=\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)=t$ and solve for $t$ :

$t=\dfrac{\sqrt{2}\pm\sqrt{10}}{4}=\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right).$

Therefore $x+\dfrac{\pi}{4}=\cdots,~\dfrac{-\pi}{10},~\dfrac{3\pi}{10},~\dfrac{11\pi}{10},$ and the maximum of $x$ occurs at $x=\dfrac{17\pi}{20}.$

$\therefore~a+b=\boxed{37}.$