Задача 5. Решите уравнение: \({\text{tg}}\,x\left( {{\text{tg}}\,x-1} \right) = 0\)
Ответ
ОТВЕТ: \(\pi k;\,\,\,\,\,\dfrac{\pi }{4} + \pi k;\quad \,k \in Z.\)
Решение
\({\rm{tg}}\,x\left( {{\rm{tg}}\,x-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\,x = 0,\,\,\,\,\,\,}\\{{\rm{tg}}\,x-1 = 0}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\,x = 0,}\\{{\rm{tg}}\,x = 1\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pi k,\,\,\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{\pi }{4} + \pi k,}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.} \right.\)
Ответ: \(\pi k;\,\,\,\,\,\dfrac{\pi }{4} + \pi k;\quad \,k \in Z.\)