Question

Помогите решить пожалуйста 4 tg x –tg 2x = 0

Answer 1

$4\, \text{tg} \, x - \text{tg} \, 2x = 0$

$4 \, \text{tg} \, x - \dfrac{2 \, \text{tg} \, x}{1 - \text{tg}^{2} \, x} = 0$

$\dfrac{4 \, \text{tg} \, x (1 - \text{tg}^{2} \, x) - 2 \, \text{tg} \, x}{1 - \text{tg}^{2} \, x} = 0$

$\dfrac{4 \, \text{tg} \, x - 4\text{tg}^{3} \, x - 2 \, \text{tg} \, x}{1 - \text{tg}^{2} \, x} = 0$

$\dfrac{2 \, \text{tg} \, x - 4 \, \text{tg}^{3} \, x}{1 - \text{tg}^{2} \, x} = 0$

$\left\{\begin{array}{ccc}2 \, \text{tg} \, x - 4 \, \text{tg}^{3} \, x = 0 \\1 - \text{tg}^{2} \, x \neq 0 \ \ \ \ \ \ \ \\\end{array}\right$

$\left\{\begin{array}{ccc}2 \, \text{tg} \, x (1 - 2 \, \text{tg}^{2} \, x) = 0\\\text{tg}^{2} \, x \neq 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\end{array}\right$

$\left\{\begin{array}{ccc} \left[\begin{array}{ccc}2\, \text{tg} \, x = 0 \ \ \ \ \ \ \\1 - 2 \, \text{tg}^{2} \, x = 0 \\\end{array}\right \ \ \ \\\ x \neq \dfrac{\pi}{4} + \dfrac{\pi n}{2}, \ n \in Z \\\end{array}\right$

$\left\{\begin{array}{ccc} \left[\begin{array}{ccc}x = \pi k, \ k \in Z \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\x = \text{arctg} \, \dfrac{\sqrt{2}}{2}+ \pi l, \ l \in Z \ \ \ \\x = -\text{arctg} \, \dfrac{\sqrt{2}}{2}+ \pi p, \ p \in Z\end{array}\right \\\ x \neq \dfrac{\pi}{4} + \dfrac{\pi n}{2}, \ n \in Z \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\end{array}\right$

Ответ: $x = \pi n; \ x = \text{arctg} \, \dfrac{\sqrt{2}}{2} + \pi k; \ x = -\text{arctg} \, \dfrac{\sqrt{2}}{2} + \pi l; \ n, \ k, \ l \in Z$