Question

т40) Решите уравнение: tg(4x+π)*tg(3x) = 1

Заранее спасибо!​

Answer 1

$\text{tg}(4x + \pi) \cdot \text{tg} (3x) = 1$

$\text{tg}(4x) \cdot \text{tg} (3x) = 1$

$\dfrac{\sin (4x)}{\cos (4x)} \cdot \dfrac{\sin (3x)}{\cos (3x)} = 1$

$\dfrac{\sin (4x) \sin (3x)}{\cos (4x) \cos (3x)} - 1 = 0$

$\dfrac{\sin (4x) \sin (3x) - \cos (4x) \cos (3x)}{\cos (4x) \cos (3x)} = 0$

$\dfrac{\cos (4x) \cos (3x) - \sin (4x) \sin (3x)}{\cos (4x) \cos (3x)} = 0$

$\dfrac{\cos (7x)}{\cos (4x) \cos (3x)} = 0$

$\left\{\begin{array}{ccc}\cos (7x) = 0\\\cos (4x) \neq 0\\\cos (3x) \neq 0\end{array}\right$

$\left\{\begin{array}{ccc}7x = \dfrac{\pi}{2} + \pi k\\4x \neq \dfrac{\pi}{2} + \pi k\\3x \neq \dfrac{\pi}{2} + \pi k\end{array}\right \ k \in Z$

$\left\{\begin{array}{ccc}x = \dfrac{\pi}{14} + \dfrac{\pi k}{7} \\x \neq \dfrac{\pi}{8} + \dfrac{\pi k}{4} \ \\x \neq \dfrac{\pi}{6} + \dfrac{\pi k}{3} \ \end{array}\right \ k \in Z$

$x = \dfrac{\pi}{14} + \dfrac{\pi k}{7}, \ k \in Z$

Ответ: $\text{B}) \ x = \dfrac{\pi}{14} + \dfrac{\pi k}{7}, \ k \in Z$