Применение основных тригонометрических формул для решения уравнений. Задача 47

ОТВЕТ: \(\dfrac{{2\pi k}}{3};\;\,\,\;\dfrac{\pi }{7} + \dfrac{{2\pi k}}{7};\;\,\,\;k \in Z.\)

\(\sin 5x-\sin 2x = 0\)

Воспользуемся формулой разности синусов:

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

\(\sin 5x-\sin 2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\sin \dfrac{{3x}}{2}\cos \dfrac{{7x}}{2} = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \dfrac{{3x}}{2} = 0,}\\{\cos \dfrac{{7x}}{2} = 0\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{{3x}}{2} = \pi k,\,\,\,\,\,\,\,\,\,}\\{\dfrac{{7x}}{2} = \dfrac{\pi }{2} + \pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{2\pi k}}{3},\,\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{\pi }{7} + \dfrac{{2\pi k}}{7},}\end{array}} \right.\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\dfrac{{2\pi k}}{3};\;\,\,\;\dfrac{\pi }{7} + \dfrac{{2\pi k}}{7};\;\,\,\;k \in Z.\)