Let $\mathrm{S}=(0,2\pi )-\left\{\frac{\pi }{2},\frac{3\pi }{4},\frac{3\pi }{2},\frac{7\pi }{4}\right\}$. Let $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right),\mathrm{x}\in \mathrm{S}$ , be the solution curve of the differential equation $\frac{dy}{dx}=\frac{1}{1+\sin 2x},y\left(\frac{\pi }{4}\right)=\frac{1}{2}$. if the sum of abscissas of all the points of intersection of the curve y=y(x) with the curve $y=\sqrt{2}\sin x$ is $\frac{k\pi }{12}$, then k is equal to

$\begin{array}{l}\frac{dy}{dx}=\frac{1}{1+\sin 2x}\\ \int dy=\int \frac{dx}{(\sin x+\cos x{)}^2}\\ \int dy=\int \frac{{\sec }^{2}x}{{\left(1+{\tan }^x\right)}^2}\\ y\left(x\right)=-\frac{1}{1+\tan x}+C\\ y\left(\frac{\pi }{4}\right)=\frac{1}{2}=-\frac{1}{2}+C\\ C=1\\ y\left(x\right)=\frac{-1}{1+\tan x}+1\\ y\left(x\right)=\frac{-1+1+\tan x}{1+\tan x}\\ y\left(x\right)=\frac{\tan x}{1+\tan x}\end{array}$

$$\begin{gathered} \mathrm{y}=\sqrt{2}\sin \mathrm{x} \\ \frac{\tan \mathrm{x}}{1+\tan \mathrm{x}}=\sqrt{2}\sin \mathrm{x} \end{gathered}$$

$\begin{array}{l}\sin x=0,\frac{1}{\sqrt{2}}=\sin x+\cos x\\ x=\pi \frac{1}{2}=\sin \left(x+\frac{\pi }{4}\right)\\ \sin \frac{\pi }{6}=\sin \left(x+\frac{\pi }{4}\right)\\ x+\frac{\pi }{4}=\pi -\frac{\pi }{6},2\pi +\frac{\pi }{6}\\ x=\frac{5\pi }{6}-\frac{\pi }{4},x=\frac{13\pi }{6}-\frac{\pi }{4}\\ x=\frac{7\pi }{12},x=\frac{23\pi }{12}\end{array}$

$\begin{array}{l}Sum of solutions=\pi +\frac{7\pi }{12}+\frac{23\pi }{12}\\ =\frac{12\pi +7\pi +23}{12}=\frac{42\pi }{12}=\frac{\mathrm{k}\pi }{12}\\ \Rightarrow \mathrm{k}=42\end{array}$