Question

решите уравнение sin6x+√3cos 6x=-2cos8x​

Answer 1

$Sin6x+\sqrt{3} Cos6x=-2Cos8x|:2\\\\\frac{\sqrt{3} }{2}Cos6x+\frac{1}{2}Sin6x=-Cos8x\\\\Cos\frac{\pi }{6}Cos6x+Sin\frac{\pi }{6} Sin6x=-Cos8x\\\\Cos(6x-\frac{\pi }{6})+Cos8x=0\\\\2Cos\frac{6x-\frac{\pi }{6}-8x }{2}Cos\frac{6x-\frac{\pi }{6}+8x }{2}=0\\\\Cos\frac{-2x-\frac{\pi }{6}} {2}Cos\frac{14x-\frac{\pi }{6}} {2}=0\\\\Cos(x+\frac{\pi }{12})Cos(7x-\frac{\pi }{12})=0\\\\1)Cos(x+\frac{\pi }{12}) =0\\\\x+\frac{\pi }{12}=\frac{\pi }{2}+\pi n,n\in Z$

$\boxed{x=\frac{5\pi }{12}+\pi n,n\in Z} \\\\2)Cos(7x-\frac{\pi }{12}) =0\\\\7x-\frac{\pi }{12}=\frac{\pi }{2}+\pi n,n\in Z\\\\7x=\frac{7\pi }{12}+\pi n,n\in Z\\\\\boxed{x=\frac{\pi }{12} +\frac{\pi n }{7} ,n \in Z}$