Применение основных тригонометрических формул для решения уравнений. Задача 45

Задача 45. Решите уравнение: \(\sin x + \sin 3x = 0\)

Ответ

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

Решение

\(\sin x + \sin 3x = 0\)

Воспользуемся формулой суммы синусов:

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

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

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

Ответ: \(\dfrac{{\pi k}}{2};\;\,\,\;k \in Z.\)