Given differential equation $\frac{dy}{dx}=\frac{1}{1+\sin 2x}$

$\int dy=\int \frac{dx}{{\left(\sin x+\cos x\right)}^2}$

$\int dy=\int \frac{{\sec }^{2}x}{{\left(1+\tan x\right)}^2}$

$y=-\frac{1}{1+\tan x}+C$

So $y\left(\frac{\pi }{4}\right)=\frac{1}{2}$

So $\frac{1}{2}=-\frac{1}{2}+C$$\Rightarrow C=1$

So $y=\frac{-1}{1+\tan x}+1$

$y=\frac{-1+1+\tan x}{1+\tan x}$

$y=\frac{\tan x}{1+\tan x}$

Solving this equation with $y=\sqrt{2}\sin x$

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

i.e. $\sin x=0, \frac{1}{\sqrt{2}}=\sin x+\cos x$

$x=\pi$

or $\frac{1}{2}=\sin \left(x+\frac{\pi }{4}\right)$

$\Rightarrow \sin \frac{\pi }{6}=\sin \left(x+\frac{\pi }{4}\right)$

$x+\frac{\pi }{4}=\pi -\frac{\pi }{6},2\pi +\frac{\pi }{6}$

$x=\frac{5\pi }{6}-\frac{\pi }{4}, x=\frac{13\pi }{6}-\frac{\pi }{4}$

$x=\frac{7\pi }{12}, x=\frac{23\pi }{12}$

Hence sum of abscissas of all the points of intersection

$=\pi +\frac{7\pi }{12}+\frac{23\pi }{12}$

$=\frac{12\pi +7\pi +23}{12}=\frac{42\pi }{12}$

$\frac{k\pi }{12}=\frac{42\pi }{12}\Rightarrow k=42$