1. Решите уравнения: 1) 2cosx = 1 2) sin(-x/3) = √2/2 3) 3 (sinx)² - 5 sin x - 2 = 0 4) 4(sinx)² - 4 cos x - 1 = 0 2. Решите неравенства: 1) cosx ≥ -1/2 2) sin(x-π/6) ≥ -1/2

1. Решите уравнения:

1) $$2\cos x = 1$$

$$\cos x = \frac{1}{2}$$

$$x = \pm \arccos \frac{1}{2} + 2\pi k, k \in \mathbb{Z}$$

$$x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}$$

Ответ: $$x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}$$

2) $$\sin(-\frac{x}{3}) = \frac{\sqrt{2}}{2}$$

$$-\frac{x}{3} = (-1)^n \arcsin \frac{\sqrt{2}}{2} + \pi n, n \in \mathbb{Z}$$

$$-\frac{x}{3} = (-1)^n \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$$

$$x = -3((-1)^n \frac{\pi}{4} + \pi n), n \in \mathbb{Z}$$

$$x = -3(-1)^n \frac{\pi}{4} - 3\pi n, n \in \mathbb{Z}$$

Ответ: $$x = -3(-1)^n \frac{\pi}{4} - 3\pi n, n \in \mathbb{Z}$$

3) $$3(\sin x)^2 - 5 \sin x - 2 = 0$$

Пусть $$t = \sin x$$, тогда

$$3t^2 - 5t - 2 = 0$$

$$D = (-5)^2 - 4 \cdot 3 \cdot (-2) = 25 + 24 = 49$$

$$t_1 = \frac{5 + \sqrt{49}}{2 \cdot 3} = \frac{5 + 7}{6} = \frac{12}{6} = 2$$

$$t_2 = \frac{5 - \sqrt{49}}{2 \cdot 3} = \frac{5 - 7}{6} = \frac{-2}{6} = -\frac{1}{3}$$

1) $$\sin x = 2$$ - нет решений, т.к. $$|\sin x| \le 1$$

2) $$\sin x = -\frac{1}{3}$$

$$x = (-1)^n \arcsin(-\frac{1}{3}) + \pi n, n \in \mathbb{Z}$$

$$x = (-1)^n (-\arcsin(\frac{1}{3})) + \pi n, n \in \mathbb{Z}$$

$$x = (-1)^{n+1} \arcsin(\frac{1}{3}) + \pi n, n \in \mathbb{Z}$$

Ответ: $$x = (-1)^{n+1} \arcsin(\frac{1}{3}) + \pi n, n \in \mathbb{Z}$$

4) $$4(\sin x)^2 - 4 \cos x - 1 = 0$$

$$4(1 - (\cos x)^2) - 4 \cos x - 1 = 0$$

$$4 - 4(\cos x)^2 - 4 \cos x - 1 = 0$$

$$-4(\cos x)^2 - 4 \cos x + 3 = 0$$

$$4(\cos x)^2 + 4 \cos x - 3 = 0$$

Пусть $$t = \cos x$$, тогда

$$4t^2 + 4t - 3 = 0$$

$$D = 4^2 - 4 \cdot 4 \cdot (-3) = 16 + 48 = 64$$

$$t_1 = \frac{-4 + \sqrt{64}}{2 \cdot 4} = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2}$$

$$t_2 = \frac{-4 - \sqrt{64}}{2 \cdot 4} = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2}$$

1) $$\cos x = \frac{1}{2}$$

$$x = \pm \arccos \frac{1}{2} + 2\pi k, k \in \mathbb{Z}$$

$$x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}$$

2) $$\cos x = -\frac{3}{2}$$ - нет решений, т.к. $$|\cos x| \le 1$$

Ответ: $$x = \pm \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}$$

2. Решите неравенства:

1) $$\cos x \ge -\frac{1}{2}$$

$$x \in [-\frac{2\pi}{3} + 2\pi k; \frac{2\pi}{3} + 2\pi k], k \in \mathbb{Z}$$

Ответ: $$x \in [-\frac{2\pi}{3} + 2\pi k; \frac{2\pi}{3} + 2\pi k], k \in \mathbb{Z}$$

2) $$\sin(x - \frac{\pi}{6}) \ge -\frac{1}{2}$$

$$x - \frac{\pi}{6} \in [-\frac{\pi}{6} + 2\pi k; \frac{7\pi}{6} + 2\pi k], k \in \mathbb{Z}$$

$$x \in [2\pi k; \frac{4\pi}{3} + 2\pi k], k \in \mathbb{Z}$$

Ответ: $$x \in [2\pi k; \frac{4\pi}{3} + 2\pi k], k \in \mathbb{Z}$$