200SX+22=0 2sinx-03=0 2 COS(-2x)=-5

Решение:

1. Первое уравнение:
   \[2\cos{x} + \sqrt{2} = 0\] \[2\cos{x} = -\sqrt{2}\] \[\cos{x} = -\frac{\sqrt{2}}{2}\] \[x = \pm \frac{3\pi}{4} + 2\pi k, k \in \mathbb{Z}\]
2. Второе уравнение:
   \[2\sin{x} - \sqrt{3} = 0\] \[2\sin{x} = \sqrt{3}\] \[\sin{x} = \frac{\sqrt{3}}{2}\] \[x = (-1)^k \frac{\pi}{3} + \pi k, k \in \mathbb{Z}\]
3. Третье уравнение:
   \[2\cos{(-2x)} = -\sqrt{3}\] \[\cos{(-2x)} = -\frac{\sqrt{3}}{2}\] \[-2x = \pm \frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z}\] \[x = \mp \frac{5\pi}{12} - \pi k, k \in \mathbb{Z}\] \[x = \mp \frac{5\pi}{12} + \pi n, n \in \mathbb{Z}\]