Solve (1+siny)dx=([2ycosy-xlogy+tany])dy

$$( 1 + \sin y ) d x = ( [ 2 y \cos y - x \log y + \tan y ] ) d y$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{2\ln(10)y\left(-y\cos(2y)-\sin(y)-y\right)}{2y\ln(y)\cos(y)+2\ln(10)\cos(y)+\ln(10)\sin(2y)}\text{, }&y\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\frac{\pi \left(2n_{1}+1\right)}{2}\text{ and }2\left(2y\log(y)\cos(y)+\sin(2y)+2\cos(y)\right)\neq 0\\x\in \mathrm{C}\text{, }&d=0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}\text{ and }y\neq 0\end{matrix}\right.$$

Solve for d

$$\left\{\begin{matrix}d=0\text{, }&y>0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}\\d\in \mathrm{R}\text{, }&y>0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\frac{\pi \left(2n_{1}+1\right)}{2}\text{ and }\frac{-4\left(y\cos(y)\right)^{2}+2xy\log(y)\cos(y)+x\sin(2y)+2x\cos(y)-2y\sin(y)}{2}=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\ln(10)y\left(2y\left(\cos(y)\right)^{2}+\sin(y)\right)}{\cos(y)\left(y\ln(y)+\ln(10)\sin(y)+\ln(10)\right)}\text{, }&y>0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}\\x\in \mathrm{R}\text{, }&d=0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}\text{ and }y>0\end{matrix}\right.$$