Задача 7. Решите уравнение: \({\text{t}}{{\text{g}}^2}x-\sqrt 3 {\text{tg}}\,x = 0\)
Ответ
ОТВЕТ: \(\pi k;\,\,\,\,\,\dfrac{\pi }{3} + \pi k;\quad \,k \in Z.\)
Решение
\({\rm{t}}{{\rm{g}}^2}x-\sqrt 3 {\rm{tg}}\,x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\rm{tg}}\,x\left( {{\rm{tg}}\,x-\sqrt 3 } \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\,x = 0,\,\,\,\,\,\,\,\,\,\,\,}\\{{\rm{tg}}\,x-\sqrt 3 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\,x = 0,\,\,\,}\\{{\rm{tg}}\,x = \sqrt 3 }\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pi k,\,\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{\pi }{3} + \pi k,}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.} \right.} \right.\)
Ответ: \(\pi k;\,\,\,\,\,\dfrac{\pi }{3} + \pi k;\quad \,k \in Z.\)