Let S = (0, 2π) - \ (π)/(2), (3π)/(4), (3π)/(2), (7π)/(4)\ . Let y...

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

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

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

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

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

So $\dfrac{1}{2} = - \dfrac{1}{2} + C$$⇒ C = 1$

So $y = \dfrac{- 1}{1 + \tan x} + 1$

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

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

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

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

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

$x = \pi$

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

$⇒ sin\dfrac{\pi}{6} = \sin \left(x + \dfrac{\pi}{4}\right)$

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

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

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

Hence sum of abscissas of all the points of intersection

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

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

$\dfrac{k\pi}{12} = \dfrac{42\pi}{12} ⇒ k = 42$