34sinx=cosx (2cos^2)x=1-sinx 2sin4x-(2sin^2)x=1 - вопрос №48366423422124221 от Madonkg 03.10.2021 17:43

Question

Алгебра: 34sinx=cosx (2cos^2)x=1-sinx 2sin4x-(2sin^2)x=1

Madonkg · Answer

1)34sinx-cosx=0/cosx34tgx-1=0tgx=1/34x=arctg1/34+πn,n∈z2)2-2sin²x-1+sinx=0sinx=a2a²-a-1=0D=1+8=9a1=(1-3)/4=-1/2⇒sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈za2=(1+3)/4=1⇒sinx=1⇒x=π/2+2πk,k∈z3)4sinxcosx-2sin²x-sin²x-cos²x=0/cos²x3tg²x-4tgx+1=0tgx=a3a²-4a+1=0D=16-12=4a1=(4-2)/6=1/3⇒tgx=1/3⇒x=arctg1/3+πn,n∈za2=(4+2)/6=1⇒tgx=1⇒x=π/4+πk,k∈z...