2sin^2 4x-4=3sin4xcos4x-4cos^2 4x

Решите задачу:

$2sin^2 4x-4=3sin4xcos4x-4cos^2 4x\\ 2sin^2 4x-4*1=3sin4xcos4x-4cos^2 4x\\ 2sin^2 4x-4sin^24x-4cos^24x=3sin4xcos4x-4cos^2 4x\\ -2sin^2 4x=3sin4xcos4x\\ 2sin^2 4x+3sin4xcos4x=0\\ sin4x(2sin4x+3cos4x)=0\\ sin4x=0 \ 2sin4x+3cos4x=0\\ 4x= \pi k, \ k \in Z\\ x= \frac{ \pi k}{4} , k \in Z\\ 2sin4x+3cos4x=0|:cos4x\\ 2tg4x+3=0\\ tg4x=-1,5\\ 4x=arctg(-1,5)+ \pi k, k \in Z\\ 4x=-arctg1,5+ \pi k\\ x=- \frac{1}{4} arctg1,5+ \frac{ \pi k}{4} , k \in Z$