Solve x_{0}=2pi/kcos2pix+x

$$x _ { 0 } = \frac { 2 \pi } { k } \cos 2 \pi x + x$$

Solve for k

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

Steps for Solving Linear Equation

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$x_{0}k=2\pi \cos(2\pi x)+kx$$

Subtract kx from both sides.

$$x_{0}k-kx=2\pi \cos(2\pi x)$$

Combine all terms containing k.

$$\left(x_{0}-x\right)k=2\pi \cos(2\pi x)$$

Divide both sides by x_{0}-x.

$$\frac{\left(x_{0}-x\right)k}{x_{0}-x}=\frac{2\pi \cos(2\pi x)}{x_{0}-x}$$

Dividing by x_{0}-x undoes the multiplication by x_{0}-x.

$$k=\frac{2\pi \cos(2\pi x)}{x_{0}-x}$$

Variable k cannot be equal to 0.

$$k=\frac{2\pi \cos(2\pi x)}{x_{0}-x}\text{, }k\neq 0$$