Задача 48. Решите уравнение: \(\cos 6x-\cos 4x = 0\)
ОТВЕТ: \(\dfrac{{\pi k}}{5};\;\,\,\;k \in Z.\)
\(\cos 6x-\cos 4x = 0\) Воспользуемся формулой разности косинусов: \(\cos \alpha -\cos \beta = -2\sin \dfrac{{\alpha + \beta }}{2}\sin \dfrac{{\alpha -\beta }}{2}.\) \(\cos 6x-\cos 4x = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,-2\sin 5x\sin x = 0\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin 5x = 0,}\\{\sin x = 0\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{5x = \pi k,}\\{x = \pi k\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \dfrac{{\pi k}}{5},}\\{x = \pi k}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x = \dfrac{{\pi k}}{5},\,\,\,\,\,k\, \in \,Z.\) Ответ: \(\dfrac{{\pi k}}{5};\;\,\,\;k \in Z.\)