t^{2} - 2t + 1 = 0\\ t_{1,2} = \frac{1 б\sqrt{4 - 4\cdot 1}}{2} = 1\\ \\ => tg(x) = 1\\ x = \frac{\pi}{4} + \pi\cdot n\\ \\ -2\pi \leq \frac{\pi}{4} + \pi\cdot n \leq \pi\\ -\frac{9}{4}\leq n \leq \frac{3}{4}\\ => n = -2,\ -1,\ 0 " alt="\frac{tg^{2}(x) + 1}{tg(x)} = 2 \ \ [-2\pi;\pi]\\ tg^{2}(x) - 2tg(x) + 1 = 0; tg(x) = t\\ => t^{2} - 2t + 1 = 0\\ t_{1,2} = \frac{1 б\sqrt{4 - 4\cdot 1}}{2} = 1\\ \\ => tg(x) = 1\\ x = \frac{\pi}{4} + \pi\cdot n\\ \\ -2\pi \leq \frac{\pi}{4} + \pi\cdot n \leq \pi\\ -\frac{9}{4}\leq n \leq \frac{3}{4}\\ => n = -2,\ -1,\ 0 " align="absmiddle" class="latex-formula">
Ответ: уравнение имеет 3 корня