Применение основных тригонометрических формул для решения уравнений. Задача 56

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

\(2{\sin ^2}x + \cos 8x = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 8x = 1-2{\sin ^2}x.\)

Воспользуемся формулой косинуса двойного угла:

\(\cos 2\alpha  = 1-2{\sin ^2}\alpha .\)

\(\cos 8x = 1-2{\sin ^2}x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 8x = \cos 2x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 8x-\cos 2x = 0.\)

Воспользуемся формулой разности косинусов:

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

\(\cos 8x-\cos 2x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-2\sin 5x\sin 3x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 5x = 0,}\\{\sin 3x = 0\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{5x = \pi k,}\\{3x = \pi k\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{\pi k}}{5},}\\{x = \dfrac{{\pi k}}{3},}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

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