2sin^2(x-п/3)+sin(x-п/3)-1=0

Решите задачу:

$2Sin^{2}(x-\frac{\pi }{3})+Sin(x-\frac{\pi }{3})-1=0\\\\Sin(x-\frac{\pi }{3})=m, -1 \leq m \leq1\\\\2m^{2}+m-1=0\\\\D=1^{2}-4*2*(-1)=1+8=9=3^{2}\\\\m_{1}=\frac{-1+3}{4}=\frac{1}{2}\\\\m_{2} =\frac{-1-3}{4}=-1$

$Sin(x-\frac{\pi }{3} )=\frac{1}{2}\\\\1)x-\frac{\pi }{3}=arcSin\frac{1}{2}+2\pi n,n\in z\\\\x-\frac{\pi }{3}=\frac{\pi }{6}+2\pi n,n\in z\\\\x=\frac{\pi }{6}+\frac{\pi }{3}+2\pi n,n\in z\\\\x_{1}=\frac{\pi }{2}+2\pi n,n\in z\\\\2)x-\frac{\pi }{3}=\pi-arcSin\frac{1}{2}+2\pi n,n\in z\\\\x-\frac{\pi }{3}=\pi-\frac{\pi }{6} +2\pi n,n\in z\\\\x-\frac{\pi }{3}=\frac{5\pi }{6}+2\pi n,n\in z\\\\x=\frac{5\pi }{6}+\frac{\pi }{3}+2\pi n,n\in z$

$x_{2}=\frac{7\pi }{6}+2\pi n,n\in z\\\\Sin(x-\frac{\pi }{3})=-1\\\\3)x-\frac{\pi }{3}=-\frac{\pi }{2}+2\pi n,n\in z\\\\x=\frac{\pi }{3}-\frac{\pi }{2}+2\pi n,n\in z\\\\x_{3}=-\frac{\pi }{6}+2\pi n,n\in z$