$\text{ If a function }y=f\left(x\right)\text{ satisfies the differential equation }$$f\left(x\right)\cdot \sin 2x-\cos x+\left(1+{\sin }^{2}x\right)f^{\mathrm{\prime }}\left(x\right)=0\text{ with initial condition }y\left(0\right)=0,\text{ then the value }$$\text{ of }f\left(\frac{\pi }{6}\right)\text{ is equal to }$

Solution:

$\begin{array}{l}\text{ The given equation is }f\left(x\right)\cdot \sin 2x-\cos x+\left(1+{\sin }^{2}x\right)f^{\mathrm{\prime }}\left(x\right)=0\\ \Rightarrow y\sin 2x-\cos x+\left(1+{\sin }^{2}x\right)\frac{dy}{dx}=0\end{array}$

$\begin{array}{l}\Rightarrow \frac{dy}{dx}+\left(\frac{\sin 2x}{1+{\sin }^{2}x}\right)y=\frac{\cos x}{1+{\sin }^{2}x}\text{ is in the form of }\frac{dy}{dx}+Py=Q\\ \text{ Where }P=\frac{\sin 2x}{1+{\sin }^{2}x}\text{ and }Q=\frac{\cos x}{1+{\sin }^{2}x}\end{array}$

$\begin{array}{l}\text{ Integrating Factor }\left(\mathrm{IF}\right)=e^{\int \frac{\sin 2x}{1+{\sin }^{2}x}dx}\\ =e^{\ln \left(1+{\sin }^{2}x\right)}\left(\because \int \frac{f^{\mathrm{\prime }}\left(x\right)}{f\left(x\right)}dx=\ln \left|f\right(x\left)\right|+\mathrm{C}\right)\\ =1+{\sin }^{2}x\\ \text{ General solution is }y\cdot \left(\mathrm{IF}\right)=\int \left(\mathrm{IF}\right)Qdx\\ \Rightarrow y\left(1+{\sin }^{2}x\right)=\int \left(1+{\sin }^{2}x\right)\frac{\cos x}{\left(1+{\sin }^{2}x\right)}dx\end{array}$

$\begin{array}{l}\Rightarrow y\left(1+{\sin }^{2}x\right)=\int \cos xdx=\sin x+C\\ \text{ When }x=0,y=0\Rightarrow C=0\\ \text{ When }x=\frac{\pi }{6},\text{ then }y\left(1+\frac{1}{4}\right)=\frac{1}{2}\\ \Rightarrow y\left(\frac{5}{4}\right)=\frac{1}{2}\\ \Rightarrow y=\frac{2}{5}\\ \therefore f\left(\frac{\pi }{6}\right)=\frac{2}{5}\end{array}$

Therefore, the correct answer is (4).