Question

Решить уравнение. ПОМОГИТЕ. С полным решением.

Answer 1

$\displaystyle sin2x+\sqrt{3}cos2x=2cos6x\bigg| :2\\\\\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x=cos6x\\\\sin\frac{\pi }{6}*sin2x+cos\frac{\pi }{6}*cos2x=cos6x\\\\cos(2x-\frac{\pi }{6})-cos6x=0\\\\-2sin(\frac{2x-\frac{\pi }{6}-6x}{2})*sin(\frac{2x-\frac{\pi }{6}+6x}{2})=0\\\\-2sin(-2x-\frac{\pi }{12})*sin(4x-\frac{\pi }{12})=0\\\\2sin(2x+\frac{\pi }{12})*sin(4x-\frac{\pi }{12})=0$

$\displaystyle sin(2x+\frac{\pi }{12})=0; 2x+\frac{\pi }{12}=\pi n; 2x=-\frac{\pi }{12}+\pi n; x=-\frac{\pi }{24}+\frac{\pi n}{2}; n\in Z \\\\sin(4x-\frac{\pi }{12})=0; 4x-\frac{\pi }{12}=\pi n; 4x=\frac{\pi }{12}+\pi n; x=\frac{\pi }{48}+\frac{\pi n}{4}; n \in Z$