|tgx+ctgx|=4/ корень из 3

$|tgx+ \frac{1}{tgx}|= \frac{4}{\sqrt{3}} \\ | \frac{ tg^{2} x+1}{tgx}|= \frac{4}{\sqrt{3}} \\ \frac{ tg^{2} x+1}{tgx}= \frac{4}{\sqrt{3}} \; \cdot|tgx \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -\frac{ tg^{2} x+1}{tgx}= \frac{4}{\sqrt{3}} \; \cdot|tgx \\ tg^{2}x+1= \frac{4 \cdot tgx}{\sqrt{3}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; -tg^{2}x-1= \frac{4 \cdot tgx}{\sqrt{3}} \\ tg^{2}x \cdot \sqrt{3} -4 \cdot tgx+ \sqrt{3}=0 \; \; \; -tg^{2}x \cdot \sqrt{3} -4 \cdot tgx- \sqrt{3}=0$

$t = tgx \\ \sqrt{3}t^{2}-4t+\sqrt{3}=0 \; \; \; \; \; \; \; \; \; \; -\sqrt{3}t^{2}-4t-\sqrt{3}=0 \\ \\ t_{1,2} = \frac{4 \pm \sqrt{16 - 12} }{2 \sqrt{3}}= \frac{4 \pm 2}{2 \sqrt{3}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; t_{3,4} = \frac{4 \pm \sqrt{16 - 12} }{-2 \sqrt{3}}= \frac{4 \pm 2}{-2 \sqrt{3}} \\ t_{1}= \frac{3}{ \sqrt{3}}=\sqrt{3}; \; } t_{2}= \frac{1}{ \sqrt{3}}= \frac{ \sqrt{3} }{3}} \; \; \; \; \; \; \;t_{3}=-\sqrt{3}; t_{4} = -\frac{ \sqrt{3} }{3}}$

$tgx = \pm \frac{ \sqrt{3} }{3}}; \; tgx= \pm \sqrt{3}\\ x_{n}= \pm \frac{\pi}{6} +\pi n, n \epsilon Z\\x_{k}= \pm \frac{\pi}{3} + \pi k, k \epsilon Z$