Question

найдите решение неравенства 1/2cos3x+√3/2sin3x<-√2/2​

Answer 1

Ответ:

$\dfrac{13\pi }{36} +\dfrac{2\pi n}{3} < x<\dfrac{19\pi }{36} +\dfrac{2\pi n}{3} ,~n\in\mathbb {Z};$

Объяснение:

$\dfrac{1}{2} cos3x+\dfrac{\sqrt{3} }{2} sin3x<-\dfrac{\sqrt{2} }{2} ;\\\\cos3x\cdot cos \dfrac{\pi }{3} +sin3x\cdot sin \dfrac{\pi }{3}<-\dfrac{\sqrt{2} }{2} ;\\\\cos( 3x-\dfrac{\pi }{3})<-\dfrac{\sqrt{2} }{2}$

$\dfrac{3\pi }{4} +2\pi n< 3x-\dfrac{\pi }{3} <\dfrac{5\pi }{4} +2\pi n,~n\in\mathbb {Z};\\\\\dfrac{3\pi }{4} +\dfrac{\pi }{3} +2\pi n< 3x<\dfrac{5\pi }{4} +\dfrac{\pi }{3} +2\pi n,~n\in\mathbb {Z};\\\\\dfrac{3\pi }{4}^{\backslash3} +\dfrac{\pi }{3}^{\backslash4} +2\pi n< 3x<\dfrac{5\pi }{4}^{\backslash3} +\dfrac{\pi }{3}^{\backslash4} +2\pi n,~n\in\mathbb {Z};$

$\dfrac{9\pi }{12} +\dfrac{4\pi }{12} +2\pi n< 3x<\dfrac{15\pi }{12} +\dfrac{4\pi }{12} +2\pi n,~n\in\mathbb {Z};$

$\dfrac{13\pi }{12} +2\pi n< 3x<\dfrac{19\pi }{12} +2\pi n,~n\in\mathbb {Z};\\\\\\\dfrac{13\pi }{36} +\dfrac{2\pi n}{3} < x<\dfrac{19\pi }{36} +\dfrac{2\pi n}{3} ,~n\in\mathbb {Z};$