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

$2sin2x-sin^2x=3cos^2x \\ \\ 2*2sinxcosx-sin^2x-3cos^2x =0 \\ \\ 4sinxcosx-sin^2x-3cos^2x=0 \\ \\ -sin^2x+4sinxcosx-3cos^2x=0 |:(-1) \\ \\ sin^2x-4sinxcosx+3cosx^2x=0 |: cos^2x \\ \\ tg^2x-4tgx+3=0 \\ \\ tgx=t \\ \\ t^2-4t+3=0 \\ \\ D=16-4*1*3=16-12=4 \\ \\ t1=(4+2)/2=6/2=3 \\ \\ t2=(4-2)/2=1$

tgx=3 tgx=1
x=arctg(3)+πn; n∈z x=π/4+πn;n∈z

Ответ:arctg(3)+πn; n∈z
π/4+πn;n∈z