Страница 183 номер 610, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (610—612). 1) \(\operatorname{tg} x = \frac{1}{\sqrt{3}}\); 2) \(\operatorname{tg} x = \sqrt{3}\); 3) \(\operatorname{tg} x = -\sqrt{3}\); 4) \(\operatorname{tg} x = -1\); 5) \(\operatorname{tg} x = 4\); 6) \(\operatorname{tg} x = -5\).
1) \(\operatorname{tg} x = \frac{1}{\sqrt{3}}\)
\(x = \operatorname{arctg} \dfrac{1}{\sqrt{3}} + \pi n,\)
\(x = \dfrac{\pi}{6} + \pi n, n \in Z;\)
2) \(\operatorname{tg} x = \sqrt{3}\)
\(x = \operatorname{arctg} \sqrt{3} + \pi n,\)
\(x = \dfrac{\pi}{3} + \pi n, n \in Z;\)
3) \(\operatorname{tg} x = - \sqrt{3}\)
\(x = - \operatorname{arctg} \sqrt{3} + \pi n,\)
\(x = - \dfrac{\pi}{3} + \pi n, n \in Z;\)
4) \(\operatorname{tg} x = - 1\)
\(x = - \operatorname{arctg} 1 + \pi n,\)
\(x = - \dfrac{\pi}{4} + \pi n, n \in Z;\)
5) \(\operatorname{tg} x = 4\)
\(x = \operatorname{arctg} 4 + \pi n, n \in Z;\)
6) \(\operatorname{tg} x = - 5\)
\(x = - \operatorname{arctg} 5 + \pi n, n \in Z.\)



