Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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