Задача 1. Решите уравнение:    \(\sin x\left( {2\sin x-1} \right) = 0\)

Ответ

ОТВЕТ: \(\pi k;\,\,\,\,\,\dfrac{\pi }{6} + 2\pi k;\quad \,\dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\) 

Решение

\(\sin x\left( {2\sin x-1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin x = 0,\,\,\,\,\,\,\,\,\,\,}\\{2\sin x-1 = 0}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\sin x = 0,}\\{\sin x = \frac{1}{2}}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\)

\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \pi k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x = \dfrac{\pi }{6} + 2\pi k,\,\,\,}\\{x = \dfrac{{5\pi }}{6} + 2\pi k,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,k\, \in \,Z.\)

Ответ:  \(\pi k;\,\,\,\,\,\dfrac{\pi }{6} + 2\pi k;\quad \,\dfrac{{5\pi }}{6} + 2\pi k;\quad k \in Z.\)