Задача 5. Решите неравенство: \({\text{tg}}\,\left( {x + \dfrac{\pi }{4}} \right) > \sqrt 3 .\)
Ответ
ОТВЕТ: \(\left( {\dfrac{\pi }{{12}} + \pi k;\,\,\dfrac{\pi }{4} + \pi k} \right),\,\,\,k \in Z.\)
Решение
\({\rm{tg}}\left( {x + \dfrac{\pi }{4}} \right) > \sqrt 3 \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{\pi }{3} + \pi k < x + \dfrac{\pi }{4} < \dfrac{\pi }{2} + \pi k\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\dfrac{\pi }{3}-\dfrac{\pi }{4} + \pi k < x < \dfrac{\pi }{2}-\dfrac{\pi }{4} + \pi k\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{\pi }{{12}} + \pi k < x < \dfrac{\pi }{4} + \pi k,\,\,\,k\, \in \,Z.\)
Ответ: \(\left( {\dfrac{\pi }{{12}} + \pi k;\dfrac{\pi }{4} + \pi k} \right),\,\,\,k\, \in \,Z.\)