Solve (2xy-sec^2x)=0

$$( 2 x y - \sec ^ { 2 } x ) = 0$$

Solve for y (complex solution)

$$y=\frac{1}{x\left(\cos(2x)+1\right)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Add \left(\sec(x)\right)^{2} to both sides.

$$2xy=0+\left(\sec(x)\right)^{2}$$

Anything plus zero gives itself.

$$2xy=\left(\sec(x)\right)^{2}$$

Divide both sides by 2x.

$$\frac{2xy}{2x}=\frac{1}{\left(\cos(x)\right)^{2}\times 2x}$$

Dividing by 2x undoes the multiplication by 2x.

$$y=\frac{1}{\left(\cos(x)\right)^{2}\times 2x}$$

Divide \frac{1}{\left(\cos(x)\right)^{2}} by 2x.

$$y=\frac{1}{2x\left(\cos(x)\right)^{2}}$$

Solve for y

$$y=\frac{1}{2x\left(\cos(x)\right)^{2}}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

Add \left(\sec(x)\right)^{2} to both sides.

$$2xy=0+\left(\sec(x)\right)^{2}$$

Anything plus zero gives itself.

$$2xy=\left(\sec(x)\right)^{2}$$

Divide both sides by 2x.

$$\frac{2xy}{2x}=\frac{1}{\left(\cos(x)\right)^{2}\times 2x}$$

Dividing by 2x undoes the multiplication by 2x.

$$y=\frac{1}{\left(\cos(x)\right)^{2}\times 2x}$$

Divide \frac{1}{\left(\cos(x)\right)^{2}} by 2x.

$$y=\frac{1}{2x\left(\cos(x)\right)^{2}}$$