Question

решите неравенства:
cosx≤√3/2
sinx≥ -√2/2
sinx≤ 0
cosx≥ -√2/2
cosx≥ -√3/2
cosx< -√3/2
cosx≤ -1/2

Answer 1

$cosx leq frac{ sqrt{3} }{2} \ frac{ pi }{6} +2 pi n leq x leq frac{11 pi }{6} +2 pi n$
$sinx geq - frac{ sqrt{2} }{2} \ - frac{ pi }{4} +2 pi n leq x leq frac{5 pi }{4} +2 pi n$
$sinx leq 0 \ pi +2 pi n leq x leq 2 pi +2 pi n$
$cosx geq - frac{ sqrt{2} }{2} \ - frac{3 pi }{4} +2 pi n leq x leq frac{3 pi }{4} +2 pi n$
$cosx geq - frac{ sqrt{3} }{2} \ - frac{5 pi }{6} +2 pi n leq x leq frac{5 pi }{6}+2 pi n$
$cosx textless - frac{ sqrt{3} }{2} \ frac{5 pi }{6}+2 pi n textless x textless frac{7 pi }{6} +2 pi n$
$cosx leq - frac{1}{2} \ frac{2 pi }{3} +2 pi n leq x leq frac{4 pi }{3} +2 pi n$