Применение основных тригонометрических формул для решения уравнений. Задача 48

Задача 48. Решите уравнение: \(\cos 6x-\cos 4x = 0\)

Ответ

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

Решение

\(\cos 6x-\cos 4x = 0\)

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