(2sinx-√2) (ctgx-√3)-0

Решим уравнение (2sinx-√2) (ctgx-√3)-0. $$2sinx-\sqrt{2} = 0$$ $$sinx = \frac{\sqrt{2}}{2}$$ $$x = arcsin(\frac{\sqrt{2}}{2}) + 2\pi n, n \in Z$$ $$x = \frac{\pi}{4} + 2\pi n, n \in Z$$ $$x = \pi - arcsin(\frac{\sqrt{2}}{2}) + 2\pi n, n \in Z$$ $$x = \pi - \frac{\pi}{4} + 2\pi n, n \in Z$$ $$x = \frac{3\pi}{4} + 2\pi n, n \in Z$$ $$ctgx-\sqrt{3} = 0$$ $$ctgx = \sqrt{3}$$ $$x = arcctg(\sqrt{3}) + \pi n, n \in Z$$ $$x = \frac{\pi}{6} + \pi n, n \in Z$$ Ответ: $$x = \frac{\pi}{4} + 2\pi n, n \in Z$$, $$x = \frac{3\pi}{4} + 2\pi n, n \in Z$$, $$x = \frac{\pi}{6} + \pi n, n \in Z$$