Question

решить уравнение:
sin2x=sin6x - cos4x

Answer 1

$sin2x=sin6x-cos4x \\ sin2x-sin6x=cos4x \\ 2sin \frac{2x-6x}{2} cos \frac{2x+6x}{2} =cos4x \\ -2sin2xcos4x=cos4x \\ -2sin2xcos4x-cos4x=0 \\ 2sin2xcos4x+cos4x=0 \\ cos4x(2sin2x+1)=0 \\ \\ cos4x=0 \\ 4x= \frac{ \pi }{2} + \pi k \\ x= \frac{ \pi }{8} + \frac{ \pi k}{4} \\ \\ 2sin2x=-1 \\ sin2x=- \frac{1}{2}$
$2x=- \frac{ \pi }{6} +2 \pi k \\ x=- \frac{ \pi }{12} +\pi k \\ \\ 2x= \frac{7 \pi }{6} +2 \pi k \\ x= \frac{7 \pi }{12} + \pi k$


Ответ: 

$\frac{ \pi }{8} + \frac{ \pi k}{4}$

$- \frac{ \pi }{12} +\pi k$

$\frac{7 \pi }{12} + \pi k$