Let y=y(x) be the solution of dy/dx+ytan x=sin 2x on [0,(π)/(2)) with y(0)=1.

The integrating factor is $e^{\int\tan x\,dx}=\sec x$, so $\frac{d}{dx}(y\sec x)=\sin2x\sec x=2\sin x$. Hence $y\sec x=-2\cos x+C$, and $y(0)=1$ gives $C=3$, so $y=3\cos x-2\cos^{2}x$. At $x=\frac\pi3$: $\frac32-\frac12=1$ (A true). Put $u=\cos x\in(0,1]$; $y=3u-2u^{2}$ is maximal at $u=\frac34$ with value $\frac94-\frac98=\frac98$ (B true), and $u=\frac34$ lies in the range. Integrating, $\int_0^{\pi/2}(3\cos x-2\cos^{2}x)dx=3-2\cdot\frac\pi4=3-\frac\pi2$ (C true). At $x=\frac\pi4$, $y=\frac{3}{\sqrt2}-1=\frac{3\sqrt2-2}{2}\approx1.12\neq1$ (D false). Answer: A, B, C. The domain restriction matters because $\tan x$ is undefined at $\frac{\pi}{2}$, but the solution extends continuously there with $y\to0$. The substitution $u=\cos x$ turns the maximisation into a simple quadratic, which is why no differentiation of the trigonometric form is needed. Option D is the typical error of evaluating $\cos\frac{\pi}{4}$ as $\frac12$ or using a wrong constant of integration.