Применение основных тригонометрических формул для решения уравнений. Задача 50

ОТВЕТ: \(\dfrac{{\pi k}}{5};\;\,\,\;-\dfrac{\pi }{6} + \dfrac{{2\pi k}}{3};\;\,\,\;k \in Z.\)

\(\cos 8x-\cos 2x = 2\sin 5x\)

Воспользуемся формулой разности косинусов:

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

\(\cos 8x-\cos 2x = 2\sin 5x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-2\sin 5x\sin 3x-2\sin 5x = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,-2\sin 5x\left( {\sin 3x + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 5x = 0,\,\,\,\,\,\,}\\{\sin 3x + 1 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 5x = 0,\,\,}\\{\sin 3x = -1}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{5x = \pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{3x = -\dfrac{\pi }{2} + 2\pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{\pi k}}{5},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = -\dfrac{\pi }{6} + \dfrac{{2\pi k}}{3},}\end{array}} \right.\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\dfrac{{\pi k}}{5};\;\,\,\;-\dfrac{\pi }{6} + \dfrac{{2\pi k}}{3};\;\,\,\;k \in Z.\)