ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 611

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\[1)\ tg\ 3x = 0\]

\[3x = arctg\ 0 + \pi n = \pi n\]

\[x = \frac{1}{3} \bullet \pi n\]

\[x = \frac{\text{πn}}{3}\]

\[Ответ:\ \ x = \frac{\text{πn}}{3}.\]

\[2)\ 1 + tg\frac{x}{3} = 0\]

\[\text{tg}\frac{x}{3} = - 1\]

\[\frac{x}{3} = arctg( - 1) + \pi n =\]

\[= - arctg\ 1 + \pi n = - \frac{\pi}{4} + \pi n\]

\[x = 3 \bullet \left( - \frac{\pi}{4} + \pi n \right)\]

\[x = - \frac{3\pi}{4} + 3\pi n\]

\[Ответ:\ x = - \frac{3\pi}{4} + 3\pi n.\]

\[3)\ \sqrt{3} + tg\frac{x}{6} = 0\]

\[\text{tg}\frac{x}{6} = - \sqrt{3}\]

\[\frac{x}{6} = arctg\left( - \sqrt{3} \right) + \pi n =\]

\[= - arctg\ \sqrt{3} + \pi n = - \frac{\pi}{3} + \pi n\]

\[x = 6 \bullet \left( - \frac{\pi}{3} + \pi n \right)\]

\[x = - 2\pi + 6\pi n\]

\[Ответ:\ x = - 2\pi + 6\pi n.\]