1) 4sin²x-2sinxcosx-3·1=0
4sin²x-2sinxcosx-3(sin²x+cos²x)=0
4sin²x-2sinxcosx-3sin²x-3cos²x=0
sin²x-2sinxcosx-3cos²x=0 |÷cos²x
tg²x-2tgx-3=0
tgx=t
t²-2t-3=0
t₁+t₂=2 t₁t₂=-3
t₁=-1 tgx=-1 x=arctg(-1)+πn x=-arctg1+πn x=-π/4+πn, n∈Z
t₂=3 tgx=3 x=arctg3+πk, k∈Z
2)3(cos²x-sin²x)+sin²x+5sinxcosx=0
3cos²x-3sin²x+sin²x+5sinxcosx=0
3cos²x-2sin²x+5sinxcosx=0 |÷cos²x
3-2tg²x+5tgx=0
tgx=t
3-2t²+5t=0
2t²-5t-3=0
D=25-4·2·(-3)=49
t₁=(5-7)/4=-1/2 tgx=-1/2 x=arctg(-1/2)+πn x=-arctg1/2+πn n∈Z
t₂=(5+7)/4=3 tgx=3 x=arctg3+πk k∈Z