$\text{ A function }y=f\left(x\right)\text{ satisfies the differential equation }f\left(x\right)·\sin 2x-\cos x+\left(1+{\sin }^{2}x\right)f^{\text{'}}\left(x\right)=0$$\text{ with initial condition }y\left(0\right)=0\text{ . The value of }f(\pi /6)=$

$y\sin 2x-\cos x+\left(1+{\sin }^{2}x\right)\frac{dy}{dx}=0$

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

$I.F=e^{\int \frac{\sin 2x}{1+{\sin }^{2}x}dx}=e^{\log \left(1+{\sin }^{2}x\right)}=1+{\sin }^{2}x$

Solution is

$y\left(1+{\sin }^{2}x\right)=\int \frac{\cos x}{\left(1+{\sin }^{2}x\right)}\left(1+{\sin }^{2}x\right)dx$

$y\left(1+{\sin }^{2}x\right)=\sin x+C\left[y\right(0)=0\Rightarrow C=0]$

$$\begin{gathered} y=\frac{\sin x}{1+{\sin }^{2}x} \\ \therefore y\left(\pi /6\right)=\frac{2}{5}=0.4 \end{gathered}$$