Тригонометрические уравнения повышенной сложности. Задача 8

ОТВЕТ: \(\dfrac{\pi }{{10}} + \dfrac{{2\pi \,k}}{5};\;\;\,\,\,k \in Z.\)

\(2{\cos ^2}x-1 = \sin 3x\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos 2x = \sin 3x\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin 3x-\sin \left( {\dfrac{\pi }{2}-2x} \right) = 0.\)

Воспользуемся формулой разности синусов:

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

\(\sin 3x-\sin \left( {\dfrac{\pi }{2}-2x} \right) = 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\sin \dfrac{{3x-\dfrac{\pi }{2} + 2x}}{2}\cos \dfrac{{3x + \dfrac{\pi }{2}-2x}}{2} = 0\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \left( {\dfrac{{5x}}{2}-\dfrac{\pi }{4}} \right) = 0,}\\{\cos \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right) = 0\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{{5x}}{2}-\dfrac{\pi }{4} = \pi k,\,\,\,\,\,\,}\\{\dfrac{x}{2} + \dfrac{\pi }{4} = \dfrac{\pi }{2} + \pi k}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{10}} + \dfrac{{2\pi k}}{5},}\\{x = \dfrac{\pi }{2} + 2\pi k,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,k \in Z.\)

Решения  \(x = \dfrac{\pi }{2} + 2\pi k,\,\,\,k\, \in \,Z\)  содержатся в решениях  \(x = \dfrac{\pi }{{10}} + \dfrac{{2\pi k}}{5},\,\,\,k \in Z.\)

Ответ:  \(\dfrac{\pi }{{10}} + \dfrac{{2\pi k}}{5},\,\,\,k\, \in Z.\)