Ответ:

Тригонометрическое неравенство .

[tex]\displaystyle \bf sin\Big(-\frac{x}{3}\Big)\leq \frac{1}{2}\\\\\\\bf -sin\frac{x}{3}\leq \frac{1}{2}\\\\\\\bf sin\frac{x}{3}\geq -\frac{1}{2}\\\\\\-\frac{\pi }{6}+2\pi n\leq \frac{x}{3}\leq \frac{7\pi }{6}+2\pi n\ \ ,\ \ n\in Z\\\\\\-\frac{\pi }{2}+6\pi n\leq x\leq \frac{7\pi }{2}+6\pi n\ \ ,\ \ n\in Z\\\\\\Otvet:\ \ x\in \Big[\, -\frac{\pi }{2}+6\pi n\ ;\ \frac{7\pi }{2}+6\pi n\ \Big]\ \ ,\ \ n\in Z\ .[/tex]