Задача 21. Решите неравенство: \({\text{t}}{{\text{g}}^2}x-\left( {1 + \sqrt 3 } \right){\text{tg}}\,x + \sqrt 3 < 0.\)
Ответ
ОТВЕТ: \(\left( {\dfrac{\pi }{4} + \pi k;\,\,\dfrac{\pi }{3} + \pi k} \right),\,\,\,k \in Z.\)
Решение
\({\rm{t}}{{\rm{g}}^2}x-\left( {1 + \sqrt 3 } \right){\rm{tg}}\,x + \sqrt 3 < 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\rm{t}}{{\rm{g}}^2}x-{\rm{tg}}\,x-\sqrt 3 {\rm{tg}}\,x + \sqrt 3 < 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,{\rm{tg}}\,x\left( {{\rm{tg}}\,x-1} \right)-\sqrt 3 \left( {{\rm{tg}}\,x-1} \right) < 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {{\rm{tg}}\,x-1} \right)\left( {{\rm{tg}}\,x-\sqrt 3 } \right) < 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,1 < {\rm{tg}}\,x < \sqrt 3 \,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {\dfrac{\pi }{4} + \pi k;\dfrac{\pi }{3} + \pi k} \right),\,\,\,\,k\, \in \,Z.\)
Ответ: \(\left( {\dfrac{\pi }{4} + \pi k;\dfrac{\pi }{3} + \pi k} \right),\,\,\,\,k\, \in \,Z.\)