If $y=y(x)$ and $\frac{2+\sin x}{y+1}(\frac{dy}{dx})=-\cos x,y(0)=1,$ then $y(\frac\pi2)=$ 
If $y=y(x)$ and $\dfrac{2+\sin x}{y+1}\left(\dfrac{dy}{dx}\right)=-\cos x,y(0)=1,$ then $y(\frac\pi2)=$

My solution:$$\frac{dy}{y+1}=-\frac{\cos x}{2+\sin x}dx\\\implies\ln|y+1|=-\ln(2+\sin x)+\ln c\\\implies|y+1|=\frac{c}{2+\sin x}$$$y(0)=1\implies c=4.$ And putting $x=\frac\pi2,$ we get, $|y+1|=\frac43\implies y=\frac13,-\frac73$
Are both the answers acceptable? The answer given in the book is only $y(\frac\pi2)=\frac13$.

Since the denominator $2+\sin x$ is always positive, the absolute value can be absorbed into the $c$
$$y + 1 = \frac{c}{2+\sin x}$$
$y(0) = 1$ fixes the branch of the function to be the positive branch, so there is only one answer.
