Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 610

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\[1)\ tg\ x = \frac{1}{\sqrt{3}}\]

\[x = arctg\ \frac{1}{\sqrt{3}} + \pi n\]

\[x = \frac{\pi}{6} + \pi n\]

\[Ответ:\ x = \ \frac{\pi}{6} + \pi n.\]

\[2)\ tg\ x = \sqrt{3}\]

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

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

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

\[3)\ tg\ x = - \sqrt{3}\]

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

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

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

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

\[4)\ tg\ x = - 1\]

\[x = arctg( - 1) + \pi n\]

\[x = - arctg\ 1 + \pi n\]

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

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

\[5)\ tg\ x = 4\]

\[x = arctg\ 4 + \pi n\]

\[Ответ:\ \ x = arctg\ 4 + \pi n.\]

\[6)\ tg\ x = - 5\]

\[x = arctg( - 5) + \pi n\]

\[x = - arctg\ 5 + \pi n\]

\[Ответ:\ x = - arctg\ 5 + \pi n.\]