Применение основных тригонометрических формул для решения уравнений. Задача 52

ОТВЕТ: \(\dfrac{\pi }{{40}} + \dfrac{{\pi k}}{{10}};\;\,\,\;\dfrac{\pi }{{20}} + \dfrac{{\pi k}}{5};\;\,\,\;k \in Z.\)

\(\cos 5x = \sin 15x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin 15x-\sin \left( {\dfrac{\pi }{2}-5x} \right) = 0.\)

Воспользуемся формулой разности синусов:

\(\sin \alpha -\sin \beta  = 2\sin \dfrac{{\alpha -\beta }}{2}\cos \dfrac{{\alpha  + \beta }}{2}.\)

\(\sin 15x-\sin \left( {\dfrac{\pi }{2}-5x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\sin \dfrac{{15x-\dfrac{\pi }{2} + 5x}}{2}\cos \dfrac{{15x + \dfrac{\pi }{2}-5x}}{2} = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \left( {10x-\dfrac{\pi }{4}} \right) = 0,}\\{\cos \left( {5x + \dfrac{\pi }{4}} \right) = 0\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{10x-\dfrac{\pi }{4} = \pi k,\,\,\,\,\,\,}\\{5x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + \pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{10x = \dfrac{\pi }{4} + \pi k,}\\{5x = \dfrac{\pi }{4} + \pi k\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{40}} + \dfrac{{\pi k}}{{10}},}\\{x = \dfrac{\pi }{{20}} + \dfrac{{\pi k}}{5},}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\dfrac{\pi }{{40}} + \dfrac{{\pi k}}{{10}};\;\,\,\;\dfrac{\pi }{{20}} + \dfrac{{\pi k}}{5};\;\,\,\;k \in Z.\)