Задача 6. Решите уравнение: \({\text{ctg}}\,x\left( {{\text{ctg}}\,x + 1} \right) = 0\)
ОТВЕТ: \(\dfrac{\pi }{2} + \pi k;\quad \,-\dfrac{\pi }{4} + \pi k;\quad k \in Z.\)
\({\rm{ctg}}\,x\left( {{\rm{ctg}}\,x + 1} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\,x = 0,\,\,\,\,\,\,}\\{{\rm{ctg}}\,x + 1 = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{ctg}}\,x = 0,\,}\\{{\rm{ctg}}\,x = -1}\end{array}} \right.} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{2} + \pi k,\,\,\,\,}\\{x = -\dfrac{\pi }{4} + \pi k,}\end{array}\,\,\,\,\,\,\,\,\,k\, \in \,Z.} \right.\) Ответ: \(\dfrac{\pi }{2} + {\rm{\pi }}k;\quad \,-\dfrac{\pi }{4} + \pi k;\quad k \in Z.\)