Задача 51. Решите уравнение: \(\sin 3x = \cos 2x\)
ОТВЕТ: \(\dfrac{\pi }{{10}} + \dfrac{{2\pi k}}{5};\;\,\,\;k \in Z.\)
\(\sin 3x = \cos 2x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sin 3x-\sin \left( {\dfrac{\pi }{2}-2x} \right) = 0.\) Воспользуемся формулой разности синусов: \(\sin \alpha -\sin \beta = 2\sin \dfrac{{\alpha -\beta }}{2}\cos \dfrac{{\alpha + \beta }}{2}.\) \(\sin 3x-\sin \left( {\dfrac{\pi }{2}-2x} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2\sin \dfrac{{3x-\dfrac{\pi }{2} + 2x}}{2}\cos \dfrac{{3x + \dfrac{\pi }{2}-2x}}{2} = 0\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin \left( {\dfrac{{5x}}{2}-\dfrac{\pi }{4}} \right) = 0,}\\{\cos \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right) = 0\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{{5x}}{2}-\dfrac{\pi }{4} = \pi k,\,\,\,\,\,\,\,}\\{\dfrac{x}{2} + \dfrac{\pi }{4} = \dfrac{\pi }{2} + \pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\dfrac{{5x}}{2} = \dfrac{\pi }{4} + \pi k,}\\{\dfrac{x}{2} = \dfrac{\pi }{4} + \pi k\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{\pi }{{10}} + \dfrac{{2\pi k}}{5},}\\{x = \dfrac{\pi }{2} + 2\pi k\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{\pi }{{10}} + \dfrac{{2\pi k}}{5},\,\,\,\,\,\,\,\,\,k\, \in \,Z.\) Ответ: \(\dfrac{\pi }{{10}} + \dfrac{{2\pi k}}{5};\;\,\,\;k \in Z.\)