f(x)=x+tanx-operatorname{ctg}x eching

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

f uchun yechish (complex solution)

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$xf=\tan(x)-\cot(x)+x$$

Ikki tarafini x ga bo‘ling.

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

x ga bo'lish x ga ko'paytirishni bekor qiladi.

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

f uchun yechish

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

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$xf=\tan(x)-\cot(x)+x$$

Ikki tarafini x ga bo‘ling.

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

x ga bo'lish x ga ko'paytirishni bekor qiladi.

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

-2\cot(2x)+x ni x ga bo'lish.

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