)20 cos2x= -1 sin 2x=1/2 sinx-√2/2=0 cosx-√2/2=0 - вопрос №49427832021212220220 от Krisitnka 22.07.2020 22:34

Question

Алгебра: )20 cos2x= -1 sin 2x=1/2 sinx-√2/2=0 cosx-√2/2=0

Krisitnka · Answer

1) 2x = πk, k∈ Z,    x = πk/2 , k∈ Z2)2x = (-1)^narcSin(1/2) + nπ, n ∈Z    2x = (-1)^n*π/6 + nπ, n ∈Z    x = (-1)^n*π/12 + nπ/2, n ∈Z3) Sinx = √2/2    x = (-1)^narcSin√2/2 + nπ, n ∈Z    x= (-1)^n*π/4 + nπ, n ∈Z4) Cosx = √2/2    x = +-arcCos√2/2 +2πk, k ∈Z    x = +- π/4 + 2πk , k ∈Z...