Question

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

Answer 1

sin^2x -3cos^2x-2sinxcosx=0
impartim toti termenii la sin^/2 x
tg^2x-2tgx-3=0
notam tgx=a
a^2-2a-3=0
a=-1
a=3
revenim la substitutie
tgx=-1 sau tgx=3
x=-π/4+kπ,k∈Z
x=arctg3+nπ,n∈Z