Question

Помогите решить уравнение:
2sin2x*sin4x+cos6x=0

Answer 1

$2\cdot sin2x\cdot sin4x+cos6x=0\\\\2\cdot sin2x\cdot (2\, sin2x\cdot cos2x)+(4cos^32x-3cos2x)=0\\\\4\, sin^22x\cdot cos2x+4cos^32x-3cos2x=0\\\\4\cdot (1-cos^22x)\cdot cos2x+4cos^32x-3cos2x=0\\\\4cos2x-4cos^32x+4cos^32x-3cos2x=0\\\\cos2x=0\\\\2x=\frac{\pi }{2}+\pi n,\; n\in Z\\\\x=\frac{\pi }{4}+\frac{\pi n}{2}\; ,\; n\in Z\\\\\star \; \; cos6x=cos(3\cdot 2x)=cos3 \alpha =4cos^3 \alpha -3cos\alpha \; ,\; \alpha =2x$

Answer 2

$2sin2x*sin4x+cos6x=0 \\ 2* \frac{1}{2}(cos2x-cos6x)+cos6x=0 \\ cos2x-cos6x+cos6x=0 \\ cos2x=0 \\ 2x= \frac{ \pi }{2}+ \pi k , k \in Z \ (:2) \\ \\ x= \frac{ \pi }{4}+ \frac{ \pi k }{2} , k \in Z$