$\begin{array}{ll} & (2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0 \\ \therefore \quad & \frac{1}{y+1} \mathrm{~d} y=\frac{-\cos x}{2+\sin x} \mathrm{~d} x \\ \therefore \quad & \text { Integrating both sides, we get } \\ & \log (y+1)=-\log (2+\sin x)+\mathrm{c} \ldots \text { (i) } \\ & \text { when } x=0, y=1 \\ & \Rightarrow \mathrm{c}=2 \log 2 \\ \therefore \quad & (\mathrm{i}) \Rightarrow \log (y+1)=-\log (2+\sin x)+\log 4 \\ & \Rightarrow \log (y+1)=\log \left(\frac{4}{2+\sin x}\right) \\ & \Rightarrow y+1=\frac{4}{2+\sin x} \\ & \quad \text { When } x=\frac{\pi}{2}, \text { (ii) } \Rightarrow y=\frac{4}{3}-1=\frac{1}{3}\end{array}$