Question

решите, пожалуйста, уравнение :(
sin6x+√3cos6x=-2cos8x

Answer 1

$\displaystyle sin6x+\sqrt{3}cos6x=-2cos8x\bigg|:2\\\\\frac{1}{2}sin6x+\frac{\sqrt{3}}{2}cos6x=-cos8x\\\\sin\frac{\pi }{6}*sin6x+cos\frac{\pi }{6}*cos6x+cos8x=0\\\\cos(6x-\frac{\pi }{6})+cos8x=0\\\\2cos(\frac{6x-\frac{\pi }{6}+8x }{2})*cos(\frac{6x-\frac{\pi }{6}-8x}{2})=0\\\\2cos(7x-\frac{\pi }{12})*cos(-x-\frac{\pi }{12})=0$

1

$\displaystyle cos(7x-\frac{\pi }{12})=0\\\\7x-\frac{\pi }{12}=\frac{\pi }{2}+\pi n; n\in Z\\\\7x=\frac{7\pi }{12}+\pi n; n\in Z\\\\x=\frac{\pi }{12}+\frac{\pi n}{7}; n\in Z$

2

$\displaystyle cos(x+\frac{\pi }{12})=0\\\\x+\frac{\pi }{12}=\frac{\pi }{2}+\pi n; n\in Z\\\\x=\frac{5\pi }{12}+\pi n; n\in Z$