Задача 20. Решите уравнение: \({\text{ct}}{{\text{g}}^2}x = \dfrac{1}{3}\)
Ответ
ОТВЕТ: \( \pm \dfrac{\pi }{3} + \pi k;\quad \,k \in Z.\)
Решение
\({\rm{ct}}{{\rm{g}}^2}x = \dfrac{1}{3}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\,x = \dfrac{1}{{\sqrt 3 }},\,\,}\\{{\rm{ctg}}\,x = -\dfrac{1}{{\sqrt 3 }}}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{3} + \pi k,\,\,\,\,}\\{x = -\dfrac{\pi }{3} + \pi k\,}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.} \right.x = \pm \dfrac{\pi }{3} + \pi k,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \( \pm \dfrac{\pi }{3} + \pi k;\quad \,k \in Z.\)