Вариант II 1. Решите неравенство: 2 sin x + 1 <0 Решите уравнения: 2. cos x + $$\frac{1}{2}$$ = 0 a) $$\pi + 2\pi n$$; B)$$(-1)^{k} \frac{\pi}{3} + \pi l$$; b) $$\pm \frac{2\pi}{3} + 2\pi n$$; г)$$\pm \frac{2\pi}{3} + 2\pi n$$; 3. tg x - $$\sqrt{3}$$ = 0 a) $$\pm \frac{2\pi}{3} + \pi n$$; B) $$\frac{\pi}{6} + \pi p$$; b) $$\frac{\pi}{3} + \pi n$$; г) $$\frac{\pi}{3} + 2\pi n$$; 4. 3 cos² x + 2 cos x - 1 = 0 a) $$\pm $$ are cos $$\frac{1}{3} + \pi n$$; Β) $$\pi + 2\pi n$$; b) $$(-1)^{l}$$ arccos$$\frac{1}{3} + n k$$; π г) +2πm; 2 5. 5 cos²x + 6 sin x - 6 = 0 a) (-1) arcsin+πk; π B)+2πm; 2 b)±arcsin+πk; Γ) π + 2πη; 6. Вычислите: 2 arc sin ($$\frac{-\sqrt{3}}{2}$$) + arc tg (-1);

1. Решим неравенство: $$2 sin x + 1 < 0$$ $$2 sin x < -1$$ $$sin x < -\frac{1}{2}$$ $$x \in \left(-\frac{5\pi}{6} + 2\pi n; -\frac{\pi}{6} + 2\pi n \right), n \in \mathbb{Z}$$

2. Решим уравнение: $$\cos x + \frac{1}{2} = 0$$ $$\cos x = -\frac{1}{2}$$ $$x = \pm \frac{2\pi}{3} + 2\pi n, n \in \mathbb{Z}$$

3. Решим уравнение: $$\text{tg } x - \sqrt{3} = 0$$ $$\text{tg } x = \sqrt{3}$$ $$x = \frac{\pi}{3} + \pi n, n \in \mathbb{Z}$$

4. Решим уравнение: $$3 \cos^2 x + 2 \cos x - 1 = 0$$ Пусть $$y = \cos x$$, тогда $$3y^2 + 2y - 1 = 0$$ $$D = 2^2 - 4 \cdot 3 \cdot (-1) = 4 + 12 = 16$$ $$y_1 = \frac{-2 + 4}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3}$$ $$y_2 = \frac{-2 - 4}{2 \cdot 3} = \frac{-6}{6} = -1$$ $$\cos x = \frac{1}{3}$$ $$x = \pm \arccos \frac{1}{3} + 2\pi n, n \in \mathbb{Z}$$ $$\cos x = -1$$ $$x = \pi + 2\pi n, n \in \mathbb{Z}$$

5. Решим уравнение: $$5 \cos^2 x + 6 \sin x - 6 = 0$$ $$5(1 - \sin^2 x) + 6 \sin x - 6 = 0$$ $$5 - 5 \sin^2 x + 6 \sin x - 6 = 0$$ $$-5 \sin^2 x + 6 \sin x - 1 = 0$$ $$5 \sin^2 x - 6 \sin x + 1 = 0$$ Пусть $$y = \sin x$$, тогда $$5y^2 - 6y + 1 = 0$$ $$D = (-6)^2 - 4 \cdot 5 \cdot 1 = 36 - 20 = 16$$ $$y_1 = \frac{6 + 4}{2 \cdot 5} = \frac{10}{10} = 1$$ $$y_2 = \frac{6 - 4}{2 \cdot 5} = \frac{2}{10} = \frac{1}{5}$$ $$\sin x = 1$$ $$x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}$$ $$\sin x = \frac{1}{5}$$ $$x = (-1)^n \arcsin \frac{1}{5} + \pi n, n \in \mathbb{Z}$$

6. Вычислим: $$2 \arcsin \left(-\frac{\sqrt{3}}{2}\right) + \text{arctg } (-1)$$ $$2 \cdot \left(-\frac{\pi}{3}\right) + \left(-\frac{\pi}{4}\right) = -\frac{2\pi}{3} - \frac{\pi}{4} = -\frac{8\pi + 3\pi}{12} = -\frac{11\pi}{12}$$

Ответ:

• 1. $$x \in \left(-\frac{5\pi}{6} + 2\pi n; -\frac{\pi}{6} + 2\pi n \right), n \in \mathbb{Z}$$
• 2. $$x = \pm \frac{2\pi}{3} + 2\pi n, n \in \mathbb{Z}$$
• 3. $$x = \frac{\pi}{3} + \pi n, n \in \mathbb{Z}$$
• 4. $$x = \pm \arccos \frac{1}{3} + 2\pi n, n \in \mathbb{Z}$$, $$x = \pi + 2\pi n, n \in \mathbb{Z}$$
• 5. $$x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}$$, $$x = (-1)^n \arcsin \frac{1}{5} + \pi n, n \in \mathbb{Z}$$
• 6. $$\frac{-11\pi}{12}$$