Решите уравнение: а) cos(π/6 - 2x) = -1; б) tg(π/4 - x/2) = -1; в) 2sin(π/3 - x/4) = √3; г) 2cos(π/4 - 3x) = √2.

**Ответ:** а) $x = \frac{7\pi}{12} + \pi n, n \in \mathbb{Z}$ б) $x = \pi + 2\pi n, n \in \mathbb{Z}$ в) $x = 4\pi n$ или $x = \frac{4\pi}{3} + 4\pi n, n \in \mathbb{Z}$ г) $x = \pi n$ или $x = \frac{\pi}{6} + \frac{2\pi n}{3}, n \in \mathbb{Z}$ **Решение:** а) $\cos\left(\frac{\pi}{6} - 2x\right) = -1$ $\frac{\pi}{6} - 2x = \pi + 2\pi n$ $-2x = \pi - \frac{\pi}{6} + 2\pi n$ $-2x = \frac{5\pi}{6} + 2\pi n$ $x = -\frac{5\pi}{12} - \pi n$ (Так как $n$ — любое целое число, знак перед $\pi n$ можно сменить на плюс, а $-\frac{5\pi}{12}$ эквивалентно $\frac{7\pi}{12}$ при сдвиге на период). б) $\text{tg}\left(\frac{\pi}{4} - \frac{x}{2}\right) = -1$ $\frac{\pi}{4} - \frac{x}{2} = -\frac{\pi}{4} + \pi n$ $-\frac{x}{2} = -\frac{\pi}{4} - \frac{\pi}{4} + \pi n$ $-\frac{x}{2} = -\frac{\pi}{2} + \pi n$ $x = \pi - 2\pi n$ в) $2\sin\left(\frac{\pi}{3} - \frac{x}{4}\right) = \sqrt{3}$ $\sin\left(\frac{\pi}{3} - \frac{x}{4}\right) = \frac{\sqrt{3}}{2}$ 1) $\frac{\pi}{3} - \frac{x}{4} = \frac{\pi}{3} + 2\pi n \Rightarrow -\frac{x}{4} = 0 + 2\pi n \Rightarrow x = -8\pi n$ 2) $\frac{\pi}{3} - \frac{x}{4} = \frac{2\pi}{3} + 2\pi n \Rightarrow -\frac{x}{4} = \frac{\pi}{3} + 2\pi n \Rightarrow x = -\frac{4\pi}{3} - 8\pi n$ г) $2\cos\left(\frac{\pi}{4} - 3x\right) = \sqrt{2}$ $\cos\left(\frac{\pi}{4} - 3x\right) = \frac{\sqrt{2}}{2}$ $\frac{\pi}{4} - 3x = \pm\frac{\pi}{4} + 2\pi n$ 1) $\frac{\pi}{4} - 3x = \frac{\pi}{4} + 2\pi n \Rightarrow -3x = 2\pi n \Rightarrow x = -\frac{2\pi n}{3}$ 2) $\frac{\pi}{4} - 3x = -\frac{\pi}{4} + 2\pi n \Rightarrow -3x = -\frac{\pi}{2} + 2\pi n \Rightarrow x = \frac{\pi}{6} - \frac{2\pi n}{3}$