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Ответ:

Решить тригонометрическое неравенство .

[tex]\bf 1)\ \ ctg(x-\dfrac{\pi }{6}) < \sqrt3[/tex]  

Сделаем замену:   [tex]\bf t=(x-\dfrac{\pi }{6})\ \ \Rightarrow \ \ \ ctg\, t < \sqrt3[/tex]

[tex]\bf \dfrac{\pi }{6}+\pi n < \ t\ < \pi +\pi n\ \ ,\ n\in Z\\\\\dfrac{\pi }{6}+\pi n < x-\dfrac{\pi }{6} < \pi +\pi n\ \ ,\ n\in Z\\\\\dfrac{\pi }{6}+\dfrac{\pi }{6}+\pi n < x < \pi +\dfrac{\pi }{6}+\pi n\ \ ,\ n\in Z\\\\\dfrac{\pi }{3}+\pi n < x < \dfrac{7\pi }{6}+\pi n\ \ ,\ n\in Z\\\\x\in \Big(\ \dfrac{\pi }{3}+\pi n\ ;\ \dfrac{7\pi }{6}+\pi n\ \Big)\ \ ,\ n\in Z[/tex]  

[tex]\displaystyle \bf 2)\ \ ctg(x+\dfrac{\pi}{3}) > \dfrac{\sqrt3}{3}\\\\Zamena:\ \ t=x+\frac{\pi }{3}\ \ \ \Rightarrow \ \ \ ctg\ t > \frac{\sqrt3}{3}\\\\0+\pi n < \ t\ < \frac{\pi }{3}+\pi n\ \ ,\ n\in Z\\\\\pi n < x+\frac{\pi }{3} < \frac{\pi }{3}+\pi n\ \ ,\ n\in Z\\\\-\frac{\pi }{3}+\pi n < x < \frac{\pi }{3}-\frac{\pi }{3}+\pi n\ \ ,\ n\in Z\\\\-\frac{\pi }{3}+\pi n < x < \pi n\ \ ,\ n\in Z\\\\x\in \Big(-\frac{\pi }{3}+\pi n\ ;\ \pi n\ \Big)\ \ ,\ n\in Z[/tex]