2.Решить уравнение: 1) cos x = -sqrt(2)/2; 2) cos x = -sqrt(3)/3; 3) sqrt(2)cos(x/4) = -1; 4) 2cos(x/3) = sqrt(3); 5) cos(2x - pi/4) = 0.

Привет! Давай решим эти тригонометрические уравнения. 1) $\cos x = -\frac{\sqrt{2}}{2}$ $x = \pm \arccos(-\frac{\sqrt{2}}{2}) + 2\pi n = \pm \frac{3\pi}{4} + 2\pi n, n \in \mathbb{Z}$ 2) $\cos x = -\frac{\sqrt{3}}{3}$ $x = \pm \arccos(-\frac{\sqrt{3}}{3}) + 2\pi n, n \in \mathbb{Z}$ 3) $\sqrt{2} \cos \frac{x}{4} = -1$ $\cos \frac{x}{4} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$ $\frac{x}{4} = \pm \frac{3\pi}{4} + 2\pi n$ $x = \pm 3\pi + 8\pi n, n \in \mathbb{Z}$ 4) $2 \cos \frac{x}{3} = \sqrt{3}$ $\cos \frac{x}{3} = \frac{\sqrt{3}}{2}$ $\frac{x}{3} = \pm \frac{\pi}{6} + 2\pi n$ $x = \pm \frac{\pi}{2} + 6\pi n, n \in \mathbb{Z}$ 5) $\cos (2x - \frac{\pi}{4}) = 0$ $2x - \frac{\pi}{4} = \frac{\pi}{2} + \pi n$ $2x = \frac{3\pi}{4} + \pi n$ $x = \frac{3\pi}{8} + \frac{\pi n}{2}, n \in \mathbb{Z}$