Уравнения, сводящиеся к простейшим заменой неизвестного. Задача 15

ОТВЕТ: \(\dfrac{\pi }{4} + \dfrac{{\pi k}}{2};\quad k \in Z.\)

\({\rm{t}}{{\rm{g}}^2}x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{\rm{tg}}\,x = 1,\,\,\,}\\{{\rm{tg}}\,x = -1}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{4} + \pi k,\,\,\,\,}\\{x = -\dfrac{\pi }{4} + \pi k\,}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.x = \dfrac{\pi }{4} + \dfrac{{\pi k}}{2},\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ:  \(\dfrac{\pi }{4} + \dfrac{{\pi k}}{2};\quad k \in Z.\)