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

1. $$\sqrt{2}\cos 2x<-1$$
2. $$2\sin 3x>-1$$
3. $$\sin\left(x+\frac{\pi}{4}\right)\leq\frac{\sqrt{2}}{2}$$
4. $$\cos\left(x-\frac{\pi}{6}\right)\geq\frac{\sqrt{3}}{2}$$

1) $$\sqrt{2}\cos 2x \le 1$$

$$\cos 2x \le \frac{1}{\sqrt{2}}$$

Для неравенства $$\cos t \le a$$ при $$a=\frac{1}{\sqrt{2}}$$ 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 неравенства $$\sin t > a$$ при $$a=-\frac12$$ имеем:

$$-\frac{\pi}{6}+2\pi n < 3x < \frac{7\pi}{6}+2\pi n,\quad n\in\mathbb Z$$

Делим на $$3$$:

$$-\frac{\pi}{18}+\frac{2\pi n}{3} < x < \frac{7\pi}{18}+\frac{2\pi n}{3},\quad n\in\mathbb Z$$

3) $$\sin\left(x+\frac{\pi}{4}\right)\le \frac{\sqrt2}{2}$$

Так как $$\frac{\sqrt2}{2}=\sin\frac{\pi}{4},$$ то

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

Вычтем $$\frac{\pi}{4}$$:

$$-\pi+2\pi n \le x \le 2\pi n,\quad n\in\mathbb Z$$

4) $$\cos\left(x-\frac{\pi}{6}\right)\ge \frac{\sqrt3}{2}$$

Так как $$\frac{\sqrt3}{2}=\cos\frac{\pi}{6},$$ то

$$-\frac{\pi}{6}+2\pi n \le x-\frac{\pi}{6} \le \frac{\pi}{6}+2\pi n,\quad n\in\mathbb Z$$

Прибавим $$\frac{\pi}{6}$$:

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

Ответ:

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

2) $$-\frac{\pi}{18}+\frac{2\pi n}{3} < x < \frac{7\pi}{18}+\frac{2\pi n}{3},\quad n\in\mathbb Z$$

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

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