Solve x^2+2xsin(x)*y+1=0

$${ x }^{ 2 } +2x \sin ( x ) \times y+1=0$$

Solve for y

$$y=-\frac{x^{2}+1}{2x\sin(x)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$2x\sin(x)y+1=-x^{2}$$

Subtract 1 from both sides.

$$2x\sin(x)y=-x^{2}-1$$

Divide both sides by 2x\sin(x).

$$\frac{2x\sin(x)y}{2x\sin(x)}=\frac{-x^{2}-1}{2x\sin(x)}$$

Dividing by 2x\sin(x) undoes the multiplication by 2x\sin(x).

$$y=\frac{-x^{2}-1}{2x\sin(x)}$$

Divide -x^{2}-1 by 2x\sin(x).

$$y=-\frac{x+\frac{1}{x}}{2\sin(x)}$$