Задача 8. Решите уравнение: \({\text{3}}\,{\text{ct}}{{\text{g}}^2}x + \sqrt 3 \,{\text{ctg}}\,x = 0\)
Ответ
ОТВЕТ: \(\dfrac{\pi }{2} + \pi k;\quad \,-\dfrac{\pi }{3} + \pi k;\quad k \in Z.\)
Решение
\(3{\rm{ct}}{{\rm{g}}^2}x + \sqrt 3 {\rm{ctg}}\,x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{ctg}}\,x\left( {3{\rm{ctg}}\,x + \sqrt 3 } \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\,x = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{3{\rm{ctg}}\,x + \sqrt 3 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\,x = 0,\,\,\,\,\,\,\,\,}\\{{\rm{ctg}}\,x = -\dfrac{{\sqrt 3 }}{3}}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{2} + \pi k,\,\,\,\,}\\{x = -\dfrac{\pi }{3} + \pi k,}\end{array}\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.} \right.} \right.\)
Ответ: \(\dfrac{\pi }{2} + \pi k;\quad \,-\dfrac{\pi }{3} + \pi k;\quad k \in Z.\)