Question

Решить неравенство 1) 2sin(x+п/6)≤√3 2) 2cos(x-п/3)> -√2 !

Answer 1

$2\sin(x+ \frac{ \pi }{6}) \leq \sqrt{3} \\\ \sin(x+ \frac{ \pi }{6}) \leq \frac{ \sqrt{3} }{2} \\\ - \frac{4 \pi }{3} +2 \pi n \leq x+ \frac{ \pi }{6} \leq \frac{ \pi }{3} +2 \pi n \\\ - \frac{8 \pi }{6} - \frac{ \pi }{6}+2 \pi n \leq x \leq \frac{2 \pi }{6}- \frac{ \pi }{6}+2 \pi n \\\ - \frac{9 \pi }{6} +2 \pi n \leq x \leq \frac{ \pi }{6}+2 \pi n \\\ - \frac{3 \pi }{2} +2 \pi n \leq x \leq \frac{ \pi }{6}+2 \pi n , \ n\in Z$

$2\cos(x- \frac{ \pi }{3} )- \sqrt{2} \\\ \cos(x- \frac{ \pi }{3} )- \frac{ \sqrt{2} }{2} \\\ - \frac{ 3\pi }{4} +2 \pi n$