Solve f(x)=x^2+tanx-operatorname{ctg}x

$$f ( x ) = x ^ { 2 } + \tan x - \operatorname { ctg } x$$

Solve for f (complex solution)

$$\left\{\begin{matrix}f=\frac{i\left(-ix^{2}\sin(2x)+2i\cos(2x)\right)}{x\sin(2x)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}\\f\in \mathrm{C}\text{, }&2\left(-ix^{2}\sin(2x)+2i\cos(2x)\right)=0\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }x=-\frac{\pi n_{2}}{2}\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$xf=\tan(x)-\cot(x)+x^{2}$$

Divide both sides by x.

$$\frac{xf}{x}=\frac{\frac{\frac{1}{\cos(x)}-2\cos(x)}{\sin(x)}+x^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$f=\frac{\frac{\frac{1}{\cos(x)}-2\cos(x)}{\sin(x)}+x^{2}}{x}$$

Divide x^{2}+\frac{\frac{1}{\cos(x)}-2\cos(x)}{\sin(x)} by x.

$$f=\frac{\frac{1}{\cos(x)}-2\cos(x)}{x\sin(x)}+x$$

Solve for f

$$f=\frac{x^{2}\sin(2x)-2\cos(2x)}{x\sin(2x)}$$

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

Steps for Solving Linear Equation

The equation is in standard form.

$$xf=\tan(x)-\cot(x)+x^{2}$$

Divide both sides by x.

$$\frac{xf}{x}=\frac{-2\cot(2x)+x^{2}}{x}$$

Dividing by x undoes the multiplication by x.

$$f=\frac{-2\cot(2x)+x^{2}}{x}$$

Divide x^{2}-2\cot(2x) by x.

$$f=-\frac{2\cot(2x)}{x}+x$$