If z = z(x) and (2 + cos x)frac{dz}{dx} + (si | vedclass

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(A) Given the differential equation: $(2 + \cos x)\frac{dz}{dx} + (\sin x)z = \sin x$. Rearranging the terms: $(2 + \cos x)\frac{dz}{dx} = \sin x - (\

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$\frac{7}{2}$

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(A) Given the differential equation: $(2 + \cos x)\frac{dz}{dx} + (\sin x)z = \sin x$. Rearranging the terms: $(2 + \cos x)\frac{dz}{dx} = \sin x - (\sin x)z = -\sin x(z - 1)$. Separating the variables: $\frac{dz}{z - 1} = \frac{-\sin x}{2 + \cos x} dx$. Integrating both sides: $\int \frac{dz}{z - 1} = \int \frac{-\sin x}{2 + \cos x} dx$. Let $u = 2 + \cos x$,then $du = -\sin x dx$. So,$\ln|z - 1| = \ln|2 + \cos x| + C$. Using the condition $z(\frac{\pi}{2}) = 3$: $\ln|3 - 1| = \ln|2 + \cos(\frac{\pi}{2})| + C \implies \ln 2 = \ln 2 + C \implies C = 0$. Thus,$z - 1 = 2 + \cos x \implies z = 3 + \cos x$. Now,find $z(\frac{\pi}{3})$: $z(\frac{\pi}{3}) = 3 + \cos(\frac{\pi}{3}) = 3 + \frac{1}{2} = \frac{7}{2}$.

If $z = z(x)$ and $(2 + \cos x)\frac{dz}{dx} + (\sin x)z = \sin x$,$z(x) > 0$ and $z(\frac{\pi}{2}) = 3$,then $z(\frac{\pi}{3})$ equals -

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