Question

2sinXcosX+3cos^2X=sin^2X

Answer 1

2sinX*cosX+3cos^2X=sin^2X

поделим на  cos^2 x

tg^2x-2tgx-3=0

tgx=3    x=arctg3+pi n. neZ

tgx=-1    x=-pi/4+pi k.  keZ