Применение основных тригонометрических формул для решения уравнений. Задача 49

ОТВЕТ: \(\dfrac{{\pi k}}{4};\;\,\,\; \pm \dfrac{\pi }{6} + \pi k;\;\,\,\;k \in Z.\)

\(\sin 2x + \sin 6x = \sin 4x\)

Воспользуемся формулой суммы синусов:

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

\(\sin 2x + \sin 6x = \sin 4x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\sin 4x\cos \left( {-2x} \right)-\sin 4x = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\sin 4x\left( {2\cos 2x-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 4x = 0,\,\,\,\,\,\,\,\,\,\,}\\{2\cos 2x-1 = 0}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 4x = 0,}\\{\cos 2x = \dfrac{1}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{4x = \pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2x =  \pm \dfrac{\pi }{3} + 2\pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{\pi k}}{4},\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x =  \pm \dfrac{\pi }{6} + \pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\dfrac{{\pi k}}{4};\;\,\,\; \pm \dfrac{\pi }{6} + \pi k;\;\,\,\;k \in Z.\)