Question

решите уравнение: cos²x+4sin²x=2sin2x

Answer 1

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