1. Решите уравнения:

Привет! Давай разберем эти тригонометрические уравнения по порядку. а) $\sin(2x + \frac{\pi}{4}) = -\frac{\sqrt{3}}{2}$ $2x + \frac{\pi}{4} = (-1)^n \arcsin(-\frac{\sqrt{3}}{2}) + \pi n, n \in \mathbb{Z}$ $2x + \frac{\pi}{4} = (-1)^n (-\frac{\pi}{3}) + \pi n$ $2x = -\frac{\pi}{4} + (-1)^{n+1} \frac{\pi}{3} + \pi n$ $x = -\frac{\pi}{8} + (-1)^{n+1} \frac{\pi}{6} + \frac{\pi n}{2}, n \in \mathbb{Z}$ б) $\cos(\frac{x}{4} - \frac{\pi}{3}) = \frac{1}{2}$ $\frac{x}{4} - \frac{\pi}{3} = \pm \frac{\pi}{3} + 2\pi n, n \in \mathbb{Z}$ $\frac{x}{4} = \frac{\pi}{3} \pm \frac{\pi}{3} + 2\pi n$ 1) $\frac{x}{4} = \frac{2\pi}{3} + 2\pi n \Rightarrow x = \frac{8\pi}{3} + 8\pi n$ 2) $\frac{x}{4} = 0 + 2\pi n \Rightarrow x = 8\pi n$ Ответ: $\frac{8\pi}{3} + 8\pi n; 8\pi n, n \in \mathbb{Z}$ в) $tg(2x + \frac{\pi}{3}) = -\frac{1}{\sqrt{3}}$ $2x + \frac{\pi}{3} = -\frac{\pi}{6} + \pi n, n \in \mathbb{Z}$ $2x = -\frac{\pi}{6} - \frac{\pi}{3} + \pi n$ $2x = -\frac{\pi}{2} + \pi n$ $x = -\frac{\pi}{4} + \frac{\pi n}{2}, n \in \mathbb{Z}$ г) $\cos 4x = -\frac{8}{3}$ Так как $-\frac{8}{3} < -1$, уравнение не имеет корней. Ответ: решений нет. д) $\sin(7x + \frac{\pi}{3}) = -1$ $7x + \frac{\pi}{3} = -\frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}$ $7x = -\frac{\pi}{2} - \frac{\pi}{3} + 2\pi n$ $7x = -\frac{5\pi}{6} + 2\pi n$ $x = -\frac{5\pi}{42} + \frac{2\pi n}{7}, n \in \mathbb{Z}$ е) $tg(2x + \frac{\pi}{6}) = 1$ $2x + \frac{\pi}{6} = \frac{\pi}{4} + \pi n, n \in \mathbb{Z}$ $2x = \frac{\pi}{4} - \frac{\pi}{6} + \pi n$ $2x = \frac{\pi}{12} + \pi n$ $x = \frac{\pi}{24} + \frac{\pi n}{2}, n \in \mathbb{Z}$ ж) $2\sin^2 x + \sin x - 1 = 0$ Пусть $\sin x = t$, где $|t| \le 1$. Получаем: $2t^2 + t - 1 = 0$ Дискриминант: $D = 1^2 - 4 \cdot 2 \cdot (-1) = 9$ $t_1 = \frac{-1 + 3}{4} = \frac{1}{2}$ $t_2 = \frac{-1 - 3}{4} = -1$ 1) $\sin x = \frac{1}{2} \Rightarrow x = (-1)^n \frac{\pi}{6} + \pi n, n \in \mathbb{Z}$ 2) $\sin x = -1 \Rightarrow x = -\frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}$ з) $\sin^2 x + \sqrt{3} \cos x \sin x = 0$ $\sin x (\sin x + \sqrt{3} \cos x) = 0$ 1) $\sin x = 0 \Rightarrow x = \pi n, n \in \mathbb{Z}$ 2) $\sin x + \sqrt{3} \cos x = 0$ Разделим на $\cos x$ (так как $\cos x \ne 0$): $tg x = -\sqrt{3} \Rightarrow x = -\frac{\pi}{3} + \pi n, n \in \mathbb{Z}$