Задача 54. Решите уравнение: \({\cos ^2}x = {\cos ^2}5x\)
ОТВЕТ: \(\dfrac{{\pi k}}{6};\;\,\,\;\dfrac{{\pi k}}{4};\;\,\,\;k \in Z.\)
\({\cos ^2}x = {\cos ^2}5x\) Воспользуемся формулой понижения степени: \({\cos ^2}\alpha = \dfrac{{1 + \cos 2\alpha }}{2}.\) \({\cos ^2}x = {\cos ^2}5x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{1 + \cos 2x}}{2} = \dfrac{{1 + \cos 10x}}{2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 10x-\cos 2x = 0.\) Воспользуемся формулой разности косинусов: \(\cos 10x-\cos 2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-2\sin 6x\sin 4x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 6x = 0,}\\{\sin 4x = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{6x = \pi k,}\\{4x = \pi k\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{\pi k}}{6},}\\{x = \dfrac{{\pi k}}{4},}\end{array}} \right.\,\,\,\,\,\,\,\,\,k\, \in \,Z.\) Ответ: \(\dfrac{{\pi k}}{6};\;\,\,\;\dfrac{{\pi k}}{4};\;\,\,\;k \in Z.\)