Solve xcos(y)+sin(2y)=xy

$$x \cos ( y ) + \sin ( 2y ) =xy$$

Solve for x (complex solution)

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

Steps for Solving Linear Equation

Subtract xy from both sides.

$$x\cos(y)+\sin(2y)-xy=0$$

Subtract \sin(2y) from both sides. Anything subtracted from zero gives its negation.

$$x\cos(y)-xy=-\sin(2y)$$

Combine all terms containing x.

$$\left(\cos(y)-y\right)x=-\sin(2y)$$

Divide both sides by \cos(y)-y.

$$\frac{\left(\cos(y)-y\right)x}{\cos(y)-y}=-\frac{\sin(2y)}{\cos(y)-y}$$

Dividing by \cos(y)-y undoes the multiplication by \cos(y)-y.

$$x=-\frac{\sin(2y)}{\cos(y)-y}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sin(2y)}{\cos(y)-y}\text{, }&\cos(y)-y\neq 0\\x\in \mathrm{R}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }y=\frac{\pi n_{1}}{2}\text{ and }\cos(y)-y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract xy from both sides.

$$x\cos(y)+\sin(2y)-xy=0$$

Subtract \sin(2y) from both sides. Anything subtracted from zero gives its negation.

$$x\cos(y)-xy=-\sin(2y)$$

Combine all terms containing x.

$$\left(\cos(y)-y\right)x=-\sin(2y)$$

Divide both sides by \cos(y)-y.

$$\frac{\left(\cos(y)-y\right)x}{\cos(y)-y}=-\frac{\sin(2y)}{\cos(y)-y}$$

Dividing by \cos(y)-y undoes the multiplication by \cos(y)-y.

$$x=-\frac{\sin(2y)}{\cos(y)-y}$$