Предмет: Математика, автор: gnomiha211

т38) Решите уравнение tg(2x+3)= tg(3x-2)

Буду благодарна, если распишете поподробнее, ибо хочу понять метод решения подобных уравнений( в моих вопросах есть еще похожие).

Заранее большое спасибо!!

Приложения:

Ответы

Автор ответа: nikebod313
1

\text{tg} (2x + 3) = \text{tg} (3x - 2)

\text{tg} (2x + 3) - \text{tg} (3x - 2) = 0

\dfrac{\sin (2x + 3)}{\cos (2x + 3)} -  \dfrac{\sin (3x - 2)}{\cos (3x - 2)} = 0

\dfrac{\sin (2x + 3)\cos(3x - 2) - \sin (3x - 2)\cos(2x + 3)}{\cos(2x + 3)\cos (3x - 2)} = 0

\dfrac{\sin (2x + 3 - 3x + 2)}{\cos(2x + 3)\cos (3x - 2)} = 0

\dfrac{\sin (-x + 5)}{\cos(2x + 3)\cos (3x - 2)} = 0

\left\{\begin{array}{ccc}\sin (-x + 5) = 0 \\\cos(2x + 3) \neq 0\\\cos(3x - 2) \neq 0\end{array}\right

\left\{\begin{array}{ccc}-x + 5 = \pi k \ \ \ \ \  \\2x + 3 \neq  \dfrac{\pi}{2} + \pi k\\3x - 2  \neq \dfrac{\pi}{2} + \pi k\end{array}\right \ k \in Z

\left\{\begin{array}{ccc}x  = 5 + \pi k \ \ \ \ \ \ \ \\x \neq \dfrac{\pi - 6}{4} + \dfrac{\pi k}{2} \\x \neq \dfrac{\pi + 4}{6} + \dfrac{\pi k}{3} \end{array}\right \ k \in Z

x = 5 + \pi k, \ k \in Z

Ответ: \text{A}) \ x = 5 + \pi k, \ k \in Z


gnomiha211: Спасибо еще раз) Вы очень помогли!
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