Упр.652 ГДЗ Алимов 10-11 класс (Алгебра)

2) 2sin3x > -1;
3) sin(x+пи/4) < =корень 2/2; 4) cos(x-пи/6) > =корень 3/2.

1. $$\sqrt{2}\cos 2x \le 1$$
   
   $$\cos 2x \le \frac{1}{\sqrt{2}}$$
   
   Для косинуса получаем:
   $$\frac{\pi}{4}+2\pi n \le 2x \le \frac{7\pi}{4}+2\pi n,\quad n\in\mathbb Z$$
   
   Делим на $2$:
   $$\frac{\pi}{8}+\pi n \le x \le \frac{7\pi}{8}+\pi n,\quad n\in\mathbb Z$$
2. $$2\sin 3x>-1$$
   
   $$\sin 3x>-\frac12$$
   
   Для синуса:
   $$-\frac{\pi}{6}+2\pi n<3x<\frac{7\pi}{6}+2\pi n,\quad n\in\mathbb Z$$
   Делим на $3$:
   $$-\frac{\pi}{18}+\frac{2\pi n}{3}
3. $$\sin\left(x+\frac{\pi}{4}\right)\le \frac{\sqrt2}{2}$$
   
   Так как $$\frac{\sqrt2}{2}=\sin\frac{\pi}{4},$$ то
   $$-\frac{3\pi}{4}+2\pi n\le x+\frac{\pi}{4}\le \frac{\pi}{4}+2\pi n,\quad n\in\mathbb Z$$
   
   Вычитаем $$\frac{\pi}{4}$$:
   $$-\pi+2\pi n\le x\le 2\pi n,\quad n\in\mathbb Z$$
4. $$\cos\left(x-\frac{\pi}{6}\right)\ge \frac{\sqrt3}{2}$$
   
   Так как $$\frac{\sqrt3}{2}=\cos\frac{\pi}{6},$$ то
   $$-\frac{\pi}{6}+2\pi n\le x-\frac{\pi}{6}\le \frac{\pi}{6}+2\pi n,\quad n\in\mathbb Z$$
   
   Прибавляем $$\frac{\pi}{6}$$:
   $$2\pi n\le x\le \frac{\pi}{3}+2\pi n,\quad n\in\mathbb Z$$

Ответ

1) $$\frac{\pi}{8}+\pi n \le x \le \frac{7\pi}{8}+\pi n,\quad n\in\mathbb Z$$

2) $$-\frac{\pi}{18}+\frac{2\pi n}{3}3) $$-\pi+2\pi n\le x\le 2\pi n,\quad n\in\mathbb Z$$

4) $$2\pi n\le x\le \frac{\pi}{3}+2\pi n,\quad n\in\mathbb Z$$