Задача 20. Решите уравнение: \(2\sin \left( {x + \frac{\pi }{4}} \right) = {\text{tg}}\;x + {\text{ctg}}\;x.\)
ОТВЕТ: \(\dfrac{\pi }{4} + 2\pi \,n,\;\;n \in Z.\)
\(2\sin \left( {x + \frac{\pi }{4}} \right) = {\rm{tg}}\,x + {\rm{ctg}}\,x\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,2\sin \left( {x + \frac{\pi }{4}} \right) = {\rm{tg}}\,x + \dfrac{1}{{{\rm{tg}}\,x}}.\) Воспользуемся тем, что \(f\left( x \right) + \dfrac{1}{{f\left( x \right)}} \ge 2\), если \(f\left( x \right) > 0\) и \(f\left( x \right) + \dfrac{1}{{f\left( x \right)}} \le — 2,\) если \(f\left( x \right) < 0\). Поэтому правая часть уравнения \({\rm{tg}}\,x{\rm{ + }}\dfrac{1}{{{\rm{tg}}\,x}}\) принимает значения \(\,\left( {-\infty ;-2} \right] \cup \left[ {2;\, + \infty } \right),\) а левая часть \(2\sin \left( {x + \dfrac{\pi }{4}} \right)\) соответственно \(\left[ {-2;\,2} \right]\). Поэтому уравнение будет иметь решение, если: \(\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{2\sin \left( {x + \dfrac{\pi }{4}} \right) = 2,}\\{{\rm{tg}}\,x + \frac{1}{{{\rm{tg}}\,x}} = 2,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{2\sin \left( {x + \dfrac{\pi }{4}} \right) = -2,}\\{{\rm{tg}}\,x + \dfrac{1}{{{\rm{tg}}\,x}} = -2\,\,\,\,\,\,\,\,\,}\end{array}} \right.\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{\sin \left( {x + \dfrac{\pi }{4}} \right) = 1,}\\{{\rm{tg}}\,x = 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{\sin \left( {x + \dfrac{\pi }{4}} \right) = -1,}\\{{\rm{tg}}\,x = -1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\end{array} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\left[ \begin{array}{l}\left\{ \begin{array}{l}x = \dfrac{{\rm{\pi }}}{4} + 2{\rm{\pi }}k,\,\,\,k \in Z\\x = \dfrac{{\rm{\pi }}}{4} + {\rm{\pi }}n,\,\,\,n \in Z\end{array} \right.\\\left\{ \begin{array}{l}x = -\dfrac{{{\rm{3\pi }}}}{4} + 2{\rm{\pi }}m,\,\,\,m \in Z\\x = -\dfrac{{\rm{\pi }}}{4} + {\rm{\pi }}p,\,\,\,p \in Z\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = \dfrac{{\rm{\pi }}}{4} + 2{\rm{\pi }}k,\,\,k \in Z.\) Ответ: \(x = \dfrac{{\rm{\pi }}}{4} + 2{\rm{\pi }}k,\,\,k \in Z.\)