Упр.743 ГДЗ Алимов 10-11 класс (Алгебра)

1) tg2x < =1; 2) tg3x < -корень 3.

1) Решим неравенство $$\tg 2x \le 1.$$

Так как $$\tg t \le 1,$$ то

$$-\frac{\pi}{2}+\pi n<t\le \frac{\pi}{4}+\pi n,\quad n\in\mathbb{Z}.$$

Положим $$t=2x$$. Тогда

$$-\frac{\pi}{2}+\pi n<2x\le \frac{\pi}{4}+\pi n,$$

$$-\frac{\pi}{4}+\frac{\pi n}{2}<x\le \frac{\pi}{8}+\frac{\pi n}{2}.$$

Отбираем решения, принадлежащие промежутку $$\left(-\frac{\pi}{2};\pi\right):$$

$$-\frac{\pi}{2}<x\le -\frac{3\pi}{8},$$

$$-\frac{\pi}{4}<x\le \frac{\pi}{8},$$

$$\frac{\pi}{4}<x\le \frac{5\pi}{8},$$

$$\frac{3\pi}{4}<x<\pi.$$

2) Решим неравенство $$\tg 3x<-\sqrt{3}.$$

Так как $$\tg t<-\sqrt{3},$$ то

$$-\frac{\pi}{2}+\pi n<t<-\frac{\pi}{3}+\pi n,\quad n\in\mathbb{Z}.$$

Положим $$t=3x$$. Тогда

$$-\frac{\pi}{2}+\pi n<3x<-\frac{\pi}{3}+\pi n,$$

$$-\frac{\pi}{6}+\frac{\pi n}{3}<x<-\frac{\pi}{9}+\frac{\pi n}{3}.$$

Отбираем решения на промежутке $$\left(-\frac{\pi}{2};\pi\right):$$

$$-\frac{\pi}{2}<x<-\frac{4\pi}{9},$$

$$-\frac{\pi}{6}<x<-\frac{\pi}{9},$$

$$\frac{\pi}{6}<x<\frac{2\pi}{9},$$

$$\frac{\pi}{2}<x<\frac{5\pi}{9},$$

$$\frac{5\pi}{6}<x<\frac{8\pi}{9}.$$

Ответ

1) $$\left(-\frac{\pi}{2};-\frac{3\pi}{8}\right]\cup\left(-\frac{\pi}{4};\frac{\pi}{8}\right]\cup\left(\frac{\pi}{4};\frac{5\pi}{8}\right]\cup\left(\frac{3\pi}{4};\pi\right);$$

2) $$\left(-\frac{\pi}{2};-\frac{4\pi}{9}\right)\cup\left(-\frac{\pi}{6};-\frac{\pi}{9}\right)\cup\left(\frac{\pi}{6};\frac{2\pi}{9}\right)\cup\left(\frac{\pi}{2};\frac{5\pi}{9}\right)\cup\left(\frac{5\pi}{6};\frac{8\pi}{9}\right).$$