\(\cos 2x\,\,\cos x = \cos 2,5x\,\,\cos 0,5x\)
Воспользуемся формулой преобразование произведение косинусов в сумму:
\(\cos \alpha \cos \beta = \dfrac{1}{2}\left( {\cos \left( {\alpha -\beta } \right) + \cos \left( {\alpha + \beta } \right)} \right).\)
\(\cos 2x\cos x = \cos 2,5x\cos 0,5x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{1}{2}\left( {\cos x + \cos 3x} \right) = \frac{1}{2}\left( {\cos 2x + \cos 3x} \right)\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\cos x = \cos 2x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos x = 2{\cos ^2}x-1.\)
Пусть \(\cos x = t,\,\,\,\,\,\,t\, \in \,\left[ {-1;1} \right]\). Тогда:
\(2{t^2}-t-1 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,\,\,\,\,}\\{t = -\dfrac{1}{2}.}\end{array}} \right.\)
Вернёмся к прежней переменной:
\(\left[ {\begin{array}{*{20}{c}}{\cos x = 1,\,\,\,\,\,}\\{\cos x = -\dfrac{1}{2}}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2\pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = \pm \dfrac{{2\pi }}{3} + 2\pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)
Ответ: \(2\pi k;\;\,\,\; \pm \dfrac{{2\pi }}{3} + 2\pi k;\,\,\,\,\,\,k \in Z.\)