Question

tg x - 2 ctg x + 1 = 0;

Answer 1

$\tan(x)-2\cot(x)+1=0$

Зная что $\cot(x)=\dfrac{1}{\tan(x)}$ подставляем в уравнение.

$\tan(x)-\dfrac{2}{\tan(x)}+1=0\\ODZ:\left \{ {{\cos(x)\neq0} \atop {\sin(x)\neq0}} \right.;\left \{ {{x\neq\frac{\pi}{2}+\pi k} \atop {x\neq}\pi k} \right., k\in\mathbb Z\Rightarrow x\neq\dfrac{\pi k}{2}, k\in\mathbb Z\\\tan^2(x)+\tan(x)-2=0\\\tan(x)=t\\t^2+t-2=0\\D=1+8=9\\t_{1,2}=-2;1\\\tan(x)=1\,\,\,and\,\,\,\tan(x)=-2\\\left \{ {{x=\frac{\pi}{4}+\pi k} \atop {x=\arctan(-2)+\pi k}} \right., k\in\mathbb Z$