Применение основных тригонометрических формул для решения уравнений. Задача 54

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