$${{dy} \over {dx}} = {1 \over {1 + \sin 2x}}$$

$$ \Rightarrow dy = {{{{\sec }^2}xdx} \over {{{(1 + \tan x)}^2}}}$$

$$ \Rightarrow y = - {1 \over {1 + \tan x}} + c$$

When $$x = {\pi \over 4}$$, $$y = {1 \over 2}$$ gives c = 1

So $$y = {{\tan x} \over {1 + \tan x}} \Rightarrow y = {{\sin x} \over {\sin x + \cos x}}$$

Now, $$y = \sqrt 2 \sin x \Rightarrow \sin x = 0$$

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

$$\sin x = 0$$ gives $$x = \pi $$ only.

and $$\sin x + \cos x = {1 \over {\sqrt 2 }} \Rightarrow \sin \left( {x + {\pi \over 4}} \right) = {1 \over 2}$$

So $$x + {\pi \over 4} = {{5\pi } \over 6}$$ or $${{13\pi } \over 6} \Rightarrow x = {{7\pi } \over {12}}$$ or $${{23\pi } \over {12}}$$

Sum of all solutions $$ = \pi + {{7\pi } \over {12}} + {{23\pi } \over {12}} = {{42\pi } \over {12}}$$

Hence, $$k = 42$$.