If the solution of the differential equation

$\frac{dy}{dx}=\frac{1}{xcosy+sin2y} is x=c{e}^{siny}-k(1+siny), then the value of k is$

$\begin{array}{l}\frac{dy}{dx}=\frac{1}{xcosy+2siny\cos y}\\ \text{}\therefore \frac{dx}{dy}=x\cos y+2\sin y\cos y\text{}\\ \frac{dx}{dy}+\left(-\cos y\right)x=2\sin y\cos y\\ \text{}I.F.=e^{-\int \cos y dx}=e^{-\sin y}\\ \text{}\therefore the solution is x.e^{-\sin y}=2\int e^{-\sin y} \sin y cosy dy\\ \text{}=-2\sin y{e}^{-\sin y}+2\int \left(e^{-\sin y}\right)\cos ydy\text{}\\ =-2\sin y{e}^{-\sin y}-2e^{-\sin y}+c\text{}\\ \Rightarrow x=-2\sin y-2+c{e}^{\sin y}\\ \text{}=c{e}^{\sin y}-2\left(1+\sin y\right)+c\\ \text{}\therefore k=2\end{array}$