Solve gcos(x-iy)/cos(x+iy)=2

$$g \frac { \cos ( x - i y ) } { \cos ( x + i y ) } = 2$$

Solve for g

$$\left\{\begin{matrix}g=\frac{2\left(e^{y+ix}+e^{3ix-y}\right)}{e^{y+3ix}+e^{ix-y}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}-iy+\frac{\pi }{2}\text{ and }e^{y+3ix}+e^{ix-y}\neq 0\\g\in \mathrm{C}\text{, }&-e^{y+ix}-e^{3ix-y}=0\text{ and }e^{y+3ix}+e^{ix-y}=0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}-iy+\frac{\pi }{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$\frac{\cos(x-iy)}{\cos(x+iy)}g=2$$

Divide both sides by \cos(x-iy)\left(\cos(x+iy)\right)^{-1}.

$$\frac{\frac{\cos(x-iy)}{\cos(x+iy)}g\cos(x+iy)}{\cos(x-iy)}=\frac{2\cos(x+iy)}{\cos(x-iy)}$$

Dividing by \cos(x-iy)\left(\cos(x+iy)\right)^{-1} undoes the multiplication by \cos(x-iy)\left(\cos(x+iy)\right)^{-1}.

$$g=\frac{2\cos(x+iy)}{\cos(x-iy)}$$