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(A) Given the differential equation: $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$. Dividing by $\cos x$,we get: $\frac{dy}{dx} + 2y 	an x = 2 \sin x$

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$\sqrt{2} - 2$

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(A) Given the differential equation: $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$. Dividing by $\cos x$,we get: $\frac{dy}{dx} + 2y 	an x = 2 \sin x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 	an x$ and $Q = 2 \sin x$. The Integrating Factor ($I$.$F$.) is $e^{\int P dx} = e^{\int 2 	an x dx} = e^{2 \ln |\sec x|} = \sec^2 x$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$. $y \sec^2 x = \int 2 \sin x \cdot \sec^2 x dx + C$. $y \sec^2 x = 2 \int 	an x \sec x dx + C$. $y \sec^2 x = 2 \sec x + C$. Given $y(\frac{\pi}{3}) = 0$,we substitute $x = \frac{\pi}{3}$ and $y = 0$: $0 \cdot \sec^2(\frac{\pi}{3}) = 2 \sec(\frac{\pi}{3}) + C$. $0 = 2(2) + C \implies C = -4$. Thus,the solution is $y \sec^2 x = 2 \sec x - 4$. For $x = \frac{\pi}{4}$,$y \sec^2(\frac{\pi}{4}) = 2 \sec(\frac{\pi}{4}) - 4$. $y(2) = 2(\sqrt{2}) - 4$. $2y = 2\sqrt{2} - 4$. $y = \sqrt{2} - 2$.

Let $y = y(x)$ be the solution of the differential equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$ for $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{3}) = 0$,then $y(\frac{\pi}{4})$ is equal to:

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