$\text{ If }y=y\left(x\right)\text{ and }\frac{2+\sin x}{y+1}\left(\frac{dy}{dx}\right)=-\cos x,y\left(0\right)=1\text{ then }y\left(\frac{\pi }{6}\right)=$

$\frac{dy}{y+1}=\frac{-\cos x}{2+\sin x}dx$

$Integrating on both sides$

$\ln \left(1+y\right)=-\log \left|2+\sin x\right|+c$

$x=0,y=1\Rightarrow \ln 2=-\ln 2+C\Rightarrow C=\ln 4$

$\ln \left(1+y\right)=-\ln \left(2+\sin x\right)+\ln 4$

$1+y=\frac{4}{2+\sin x}$

$putx=\frac{\pi }{6}\Rightarrow 1+y=\frac{4}{2+\frac{1}{2}}=\frac{8}{5}\Rightarrow y=\frac{8}{5}-1=\frac{3}{5}$