Let S = (0, 2 pi) - left{frac{pi}{2}, frac{3 | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = \frac{1}{1 + \sin 2x}$.Integrating both sides: $\int dy = \int \frac{dx}{(\sin x + \cos x)^2} = \

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$42$

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(D) Given the differential equation $\frac{dy}{dx} = \frac{1}{1 + \sin 2x}$.Integrating both sides: $\int dy = \int \frac{dx}{(\sin x + \cos x)^2} = \int \frac{\sec^2 x}{(1 + 	an x)^2} dx$.Let $u = 1 + 	an x$,then $du = \sec^2 x dx$.So,$y = \int u^{-2} du = -u^{-1} + C = -\frac{1}{1 + 	an x} + C$.Using the condition $y\left(\frac{\pi}{4}ight) = \frac{1}{2}$,we get $\frac{1}{2} = -\frac{1}{1 + 1} + C \Rightarrow \frac{1}{2} = -\frac{1}{2} + C \Rightarrow C = 1$.Thus,$y(x) = 1 - \frac{1}{1 + 	an x} = \frac{	an x}{1 + 	an x}$.Now,equate $y(x)$ with $y = \sqrt{2} \sin x$: $\frac{	an x}{1 + 	an x} = \sqrt{2} \sin x$.$\frac{\sin x}{\cos x + \sin x} = \sqrt{2} \sin x$.This gives $\sin x = 0$ (so $x = \pi$ in $(0, 2\pi)$) or $\frac{1}{\cos x + \sin x} = \sqrt{2} \Rightarrow \sin x + \cos x = \frac{1}{\sqrt{2}}$.$\sqrt{2} \sin\left(x + \frac{\pi}{4}ight) = \frac{1}{\sqrt{2}} \Rightarrow \sin\left(x + \frac{\pi}{4}ight) = \frac{1}{2}$.$x + \frac{\pi}{4} = \frac{\pi}{6} + 2n\pi$ or $x + \frac{\pi}{4} = \pi - \frac{\pi}{6} + 2n\pi$.For $x \in (0, 2\pi)$,$x + \frac{\pi}{4} = \frac{5\pi}{6}$ or $x + \frac{\pi}{4} = \frac{13\pi}{6}$.$x = \frac{5\pi}{6} - \frac{\pi}{4} = \frac{7\pi}{12}$ and $x = \frac{13\pi}{6} - \frac{\pi}{4} = \frac{23\pi}{12}$.Sum of abscissas $= \pi + \frac{7\pi}{12} + \frac{23\pi}{12} = \frac{12\pi + 7\pi + 23\pi}{12} = \frac{42\pi}{12}$.Comparing with $\frac{k\pi}{12}$,we get $k = 42$.

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