Задача 34. Решите уравнение: \({\text{t}}{{\text{g}}^2}2x + 2\sqrt 3 \;{\text{tg}}\,2x + 3 = -{\text{ct}}{{\text{g}}^2}\left( {4y-\dfrac{\pi }{6}} \right).\)
Ответ
ОТВЕТ: \(\left( {\;-\dfrac{\pi }{6} + \dfrac{{\pi \,n}}{2};\;\dfrac{\pi }{6} + \dfrac{{\pi \,k}}{4}} \right),\;\;n,k \in Z.\)
Решение
\({\rm{t}}{{\rm{g}}^2}2x + 2\sqrt 3 {\rm{tg}}\,2x + 3 = -{\rm{ct}}{{\rm{g}}^2}\left( {4y-\dfrac{\pi }{6}} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{\left( {{\rm{tg}}\,2x + \sqrt 3 } \right)^2}{\rm{ + ct}}{{\rm{g}}^2}\left( {4y-\dfrac{\pi }{6}} \right) = 0.\)
Полученное уравнение будет иметь решение, если:
\(\left\{ {\begin{array}{*{20}{c}}{{\rm{tg}}2x + \sqrt 3 = 0,\,\,\,\,\,}\\{{\rm{ctg}}\left( {4y-\dfrac{\pi }{6}} \right) = 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}2x = -\dfrac{{\rm{\pi }}}{3} + {\rm{\pi }}n,\\4y-\dfrac{\pi }{6} = \dfrac{{\rm{\pi }}}{2} + {\rm{\pi }}k\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = -\dfrac{{\rm{\pi }}}{6} + \dfrac{{{\rm{\pi }}n}}{2},\\y = \dfrac{{\rm{\pi }}}{6} + \dfrac{{{\rm{\pi }}k}}{4}\end{array} \right.\,\,\,\,n,k \in Z.\)
Ответ: \(\left( {\;-\dfrac{{\rm{\pi }}}{6} + \dfrac{{{\rm{\pi }}\,n}}{2};\;\dfrac{{\rm{\pi }}}{6} + \dfrac{{{\rm{\pi }}\,k}}{4}} \right),\;\;n,k \in Z.\)