Solve f(x)/x=cos(x-3)

$$\frac { f ( x ) } { x } = \cos ( x - 3 )$$

Solve for f (complex solution)

$$f=\frac{e^{ix-3i}+e^{-ix+3i}}{2}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$fx=x\cos(x-3)$$

The equation is in standard form.

$$xf=x\cos(x-3)$$

Divide both sides by x.

$$\frac{xf}{x}=\frac{x\cos(x-3)}{x}$$

Dividing by x undoes the multiplication by x.

$$f=\frac{x\cos(x-3)}{x}$$

Divide x\cos(x-3) by x.

$$f=\cos(x-3)$$

Solve for f

$$f=\cos(x-3)$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$fx=x\cos(x-3)$$

The equation is in standard form.

$$xf=x\cos(x-3)$$

Divide both sides by x.

$$\frac{xf}{x}=\frac{x\cos(x-3)}{x}$$

Dividing by x undoes the multiplication by x.

$$f=\frac{x\cos(x-3)}{x}$$

Divide x\cos(x-3) by x.

$$f=\cos(x-3)$$

Solve for x (complex solution)

$$x=2\pi n_{1}+3+\left(-i\right)\ln(f+\left(-1\right)\left(f^{2}-1\right)^{\frac{1}{2}})\text{, }n_{1}\in \mathrm{Z}\text{, }\left(not(n_{1}=\left(-\frac{1}{2}\right)\left(3+\left(-i\right)\ln(f+\left(-1\right)\left(f^{2}-1\right)^{\frac{1}{2}})\right)\pi ^{-1})\text{ and }not(n_{1}=\left(-\frac{3}{2}+\frac{1}{2}i\ln(f+\left(-1\right)\left(f^{2}-1\right)^{\frac{1}{2}})\right)\pi ^{-1})\right)\text{ and }not(2\pi n_{1}+3+\left(-i\right)\ln(f+\left(-1\right)\left(f^{2}-1\right)^{\frac{1}{2}})=0)$$

$$x=2\pi n_{2}+3+\left(-i\right)\ln(f+\left(f^{2}-1\right)^{\frac{1}{2}})\text{, }n_{2}\in \mathrm{Z}\text{, }not(2\pi n_{2}+3+\left(-i\right)\ln(f+\left(f^{2}-1\right)^{\frac{1}{2}})=0)\text{ and }not(2\pi n_{2}+3+\left(-i\right)\ln(f+\left(f^{2}-1\right)^{\frac{1}{2}})=0)$$