Question

помогите :')
$\sin(x) \leqslant \frac{1}{2}$
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Answer 1

$\displaystyle\bf\\Sinx\leq \frac{1}{2} \\\\\\-\pi -arcSin\frac{1}{2}+2\pi n \leq x\leq arcSin\frac{1}{2} +2\pi n \ , \ n\in Z\\\\\\-\pi -\frac{\pi }{6} +2\pi n\leq x\leq \frac{\pi }{6} +2\pi n \ , \ n\in Z\\\\\\-\frac{7\pi }{6}+2\pi n\leq x\leq \frac{\pi }{6} +2\pi n \ , \ n\in Z\\\\\\Otvet:x\in\Big[-\frac{7\pi }{6} +2\pi n \ ; \ \frac{\pi }{6} +2\pi n\Big] \ , \ n\in Z$