Задача 2. Решите неравенство: \(\sin \left( {2x + \dfrac{\pi }{3}} \right) \leqslant \frac{1}{2}.\)
Ответ
ОТВЕТ: \(\left[ {\dfrac{\pi }{4} + \pi k;\,\,\dfrac{{11\pi }}{{12}} + \pi k} \right],\,\,\,k \in Z.\)
Решение
\(\sin \left( {2x + \dfrac{\pi }{3}} \right) \le \dfrac{1}{2}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{5\pi }}{6} + 2\pi k \le 2x + \dfrac{\pi }{3} \le \dfrac{{13\pi }}{6} + 2\pi k\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\dfrac{{5\pi }}{6}-\dfrac{\pi }{3} + 2\pi k \le 2x \le \dfrac{{13\pi }}{6}-\dfrac{\pi }{3} + 2\pi k\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{\pi }{2} + 2\pi k \le 2x \le \dfrac{{11\pi }}{6} + 2\pi k\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\dfrac{\pi }{4} + \pi k \le x \le \dfrac{{11\pi }}{{12}} + \pi k,\,\,\,k\, \in \,Z.\)
Ответ: \(\left[ {\dfrac{\pi }{4} + \pi k;\dfrac{{11\pi }}{{12}} + \pi k} \right],\,\,\,k\, \in \,Z.\)