9sinxcosx-7 cos2x=2sin^2x

Вроде так..2sin²x-9sinxcosx+7cos²x=0 (:cos²x≠0) 2tg²x-9tgx+7=0 tgx=y;2y²-9y+7=0;y₁=1;y₂=7/2 1) tgx=1;x=π/4+πn,n∈Z 2) tgx=7/2;x=arctg(7/2)+πk,k∈Z.