Question

2sin2x - cosx = √3sinx

Answer 1

Ответ:

$x=\frac{\pi }{6}+2\pi n,n \in Z\\ \\x=\frac{5\pi }{18}+\frac{2}{3} \pi n,n \in Z$

Пошаговое объяснение:

$2sin2x - cosx = \sqrt{3} sinx \; |:2\\ \\ sin2x-(\frac{1}{2}cosx+\frac{\sqrt{3} }{2}sinx)=0\\ \\ sin\frac{\pi }{6}=\frac{1}{2}\\ \\ cos\frac{\pi }{6} =\frac{1}{2} \\ \\ sin(\alpha +\beta )=sin\alpha *cos\beta +cos\alpha *sin\beta$

$sin2x-(sin\frac{\pi }{6}*cosx+cos\frac{\pi }{6}*sinx)=0\\ \\ sin2x-sin(x+\frac{\pi }{6} )=0$

$sin\alpha -sin\beta =2sin\frac{\alpha -\beta }{2}*cos\frac{\alpha +\beta }{2}$

$2sin\frac{2x-x-\frac{\pi }{6} }{2}*cos\frac{2x+x+\frac{\pi }{6} }{2}=0\\ \\ sin(\frac{x}{2}-\frac{\pi }{12})* cos(\frac{3x}{2}+\frac{\pi }{12} )=0\\ \\ sin(\frac{x}{2}-\frac{\pi }{12})=0\\ \\ \frac{x}{2}-\frac{\pi }{12}=\pi n,n \in Z\\ \\ \frac{x}{2}=\frac{\pi }{12}+ \pi n ,n \in Z\\ \\ x=\frac{\pi }{6}+2\pi n,n \in Z$

$cos(\frac{3x}{2}+\frac{\pi }{12})=0\\ \\ \frac{3x}{2}+\frac{\pi }{12}=\frac{\pi }{2}+\pi n,n \in Z\\ \\ \frac{3x}{2}=\frac{5\pi }{12}+\pi n,n \in Z\\ \\ x=\frac{5\pi }{18}+\frac{2}{3} \pi n,n \in Z$