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(D) The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.Rearranging the terms to separate the variables,we get $\frac

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$\frac{\pi}{2}$

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(D) The given differential equation is $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.Rearranging the terms to separate the variables,we get $\frac{dy}{y+1} = -\frac{\cos x}{1+\sin x} dx$.Integrating both sides,we have $\int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} dx$.This yields $\ln|y+1| = -\ln|1+\sin x| + C$,which simplifies to $\ln|y+1| + \ln|1+\sin x| = C$.Using the property of logarithms,we get $(y+1)(1+\sin x) = K$,where $K = e^C$.Given the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$ into the equation: $(0+1)(1+\sin 0) = K$,which gives $1(1+0) = K$,so $K = 1$.Thus,the particular solution is $(y+1)(1+\sin x) = 1$.If the curve passes through $(\alpha, -\frac{1}{2})$,we substitute $x = \alpha$ and $y = -\frac{1}{2}$ into the equation:$(-\frac{1}{2} + 1)(1+\sin \alpha) = 1$.$\frac{1}{2}(1+\sin \alpha) = 1$.$1+\sin \alpha = 2$.$\sin \alpha = 1$.Therefore,$\alpha = \frac{\pi}{2}$.

Let $y = y(x)$ be the solution curve of the differential equation $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$ with the condition $y(0) = 0$. If the curve $y = y(x)$ passes through the point $(\alpha, -\frac{1}{2})$,then a value of $\alpha$ is:

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