1. Вычислить: а) arccos 0 + arctg 3 .

### 1. Вычислить: a) $\arccos 0 + \text{arctg } \sqrt{3} = \frac{\pi}{2} + \frac{\pi}{3} = \frac{3\pi+2\pi}{6} = \frac{5\pi}{6}$ б) $\arcsin\left(-\frac{1}{2}\right) + \arccos\frac{1}{2} = -\frac{\pi}{6} + \frac{\pi}{3} = \frac{-\pi + 2\pi}{6} = \frac{\pi}{6}$ в) $2\arcsin\frac{1}{2} + 3\arccos\frac{\sqrt{3}}{2} = 2\cdot\frac{\pi}{6} + 3\cdot\frac{\pi}{6} = \frac{2\pi}{6} + \frac{3\pi}{6} = \frac{5\pi}{6}$ ### 2. Решить уравнения: a) $2\sin 2x - \sqrt{2} = 0 \implies \sin 2x = \frac{\sqrt{2}}{2}$ $2x = (-1)^n \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$ $x = (-1)^n \frac{\pi}{8} + \frac{\pi n}{2}, n \in \mathbb{Z}$ б) $\cos\left(\frac{x}{2} - \frac{\pi}{7}\right) = \frac{\sqrt{3}}{2}$ $\frac{x}{2} - \frac{\pi}{7} = \pm\frac{\pi}{6} + 2\pi n, n \in \mathbb{Z}$ $\frac{x}{2} = \frac{\pi}{7} \pm \frac{\pi}{6} + 2\pi n$ $x = \frac{2\pi}{7} \pm \frac{\pi}{3} + 4\pi n, n \in \mathbb{Z}$ в) $\text{tg } 2x - \sqrt{3} = 0 \implies \text{tg } 2x = \sqrt{3}$ $2x = \frac{\pi}{3} + \pi n, n \in \mathbb{Z}$ $x = \frac{\pi}{6} + \frac{\pi n}{2}, n \in \mathbb{Z}$ г) $\cos^2 x + 6\sin x - 6 = 0$ $1 - \sin^2 x + 6\sin x - 6 = 0 \implies -\sin^2 x + 6\sin x - 5 = 0$ $\sin^2 x - 6\sin x + 5 = 0$ Пусть $\sin x = t$, где $|t| \le 1$ $t^2 - 6t + 5 = 0$. По теореме Виета $t_1 = 1, t_2 = 5$ (не подходит, т.к. $>1$) $\sin x = 1 \implies x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}$ д) $3\sin^2 x - 8\sin x \cos x + 5\cos^2 x = 0$ Разделим на $\cos^2 x$ (так как $\cos x \neq 0$): $3\text{tg}^2 x - 8\text{tg } x + 5 = 0$ Пусть $\text{tg } x = t$: $3t^2 - 8t + 5 = 0$. Корни $t_1 = 1, t_2 = \frac{5}{3}$ 1) $\text{tg } x = 1 \implies x = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$ 2) $\text{tg } x = \frac{5}{3} \implies x = \text{arctg } \frac{5}{3} + \pi n, n \in \mathbb{Z}$ е) $\cos 5x + \cos x = 0$ $2\cos\frac{5x+x}{2} \cos\frac{5x-x}{2} = 0$ $2\cos 3x \cos 2x = 0$ 1) $\cos 3x = 0 \implies 3x = \frac{\pi}{2} + \pi n \implies x = \frac{\pi}{6} + \frac{\pi n}{3}, n \in \mathbb{Z}$ 2) $\cos 2x = 0 \implies 2x = \frac{\pi}{2} + \pi k \implies x = \frac{\pi}{4} + \frac{\pi k}{2}, k \in \mathbb{Z}$