Solve z=ycos2x

$$z = y \cos 2 x$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=\frac{z}{\cos(2x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\\y\in \mathrm{C}\text{, }&z=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$y\cos(2x)=z$$

The equation is in standard form.

$$\cos(2x)y=z$$

Divide both sides by \cos(2x).

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

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

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

Solve for y

$$\left\{\begin{matrix}y=\frac{z}{\cos(2x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\\y\in \mathrm{R}\text{, }&z=0\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$y\cos(2x)=z$$

The equation is in standard form.

$$\cos(2x)y=z$$

Divide both sides by \cos(2x).

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

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

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\pi n_{1}-\frac{i\ln(\frac{-\sqrt{z^{2}-y^{2}}+z}{y})}{2}\text{, }n_{1}\in \mathrm{Z}\text{; }x=\pi n_{2}-\frac{i\ln(\frac{\sqrt{z^{2}-y^{2}}+z}{y})}{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&y\neq 0\\x\in \mathrm{C}\text{, }&z=0\text{ and }y=0\end{matrix}\right.$$