Применение основных тригонометрических формул для решения уравнений. Задача 57

ОТВЕТ: \(\dfrac{\pi }{{10}} + \dfrac{{\pi k}}{5};\;\,\,\;\dfrac{\pi }{6} + \dfrac{{\pi k}}{3};\,\,\,\,\,\,k \in Z.\)

\(2{\sin ^2}4x-1 = {\cos ^2}x-{\sin ^2}x\)

Воспользуемся формулами косинуса двойного угла:

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

\(2{\sin ^2}4x-1 = {\cos ^2}x-{\sin ^2}x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 2x + 1-2{\sin ^2}4x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 2x + \cos 8x = 0.\)

Воспользуемся формулой суммы косинусов:

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

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

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{5x = \dfrac{\pi }{2} + \pi k,}\\{3x = \dfrac{\pi }{2} + \pi k}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{10}} + \dfrac{{\pi k}}{5},}\\{x = \dfrac{\pi }{6} + \dfrac{{\pi k}}{3},}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ: \(\dfrac{\pi }{{10}} + \dfrac{{\pi k}}{5};\;\,\,\;\dfrac{\pi }{6} + \dfrac{{\pi k}}{3};\,\,\,\,\,\,k \in Z.\)