Question

cos^2(x)+4sin^2(x)=2sin(2x)

Answer 1

$cos^2x+4sin^2x=2sin2x\ \ cos^2x+4sin^2x=4sin xcos x|:cos^2x\ \ 1+4tg^2x=4tgx\ \ 4tg^2x-4tg x+1=0\ \(2tgx-1)^2=0\ \ 2tg x-1=0\ \ tgx=0.5\ \ x=arctg0.5+ pi n,n in Z$