Question

решить неравенства 2sin x +√3 <0​

Answer 1

$2Sinx+\sqrt{3} <0\\\\2Sinx<-\sqrt{3}\\\\Sinx<-\dfrac{\sqrt{3} }{2} \\\\-\pi-arc Sin\Big(-\dfrac{\sqrt{3} }{2}\Big)+2\pi n<x<arc Sin\Big(-\dfrac{\sqrt{3} }{2}\Big)+2\pi n,n\in Z\\\\-\pi+\dfrac{\pi }{3}+2\pi n<x< -\dfrac{\pi }{3}+2\pi n,n\in Z\\\\-\dfrac{2\pi }{3}+2\pi n<x<-\dfrac{\pi }{3}+2\pi n,n\in Z \\\\Otvet:\boxed{x\in\Big(-\dfrac{2\pi }{3}+2\pi n \ ; \ -\dfrac{\pi }{3}+2\pi n\Big),n\in Z}$