разложения на множители: 4sin⁴(x-π)+cos²(3π/2 -2x)=2 - вопрос №8138631432225232 от Werty0183 23.01.2020 01:05

Question

Алгебра: разложения на множители: 4sin⁴(x-π)+cos²(3π/2 -2x)=2 [-5π/2; -3π/2]

Werty0183 · Answer

4Sin⁴(x-π) + Cos²(3π/2-2x) = 24Sin⁴x + Sin²2x = 24Sin⁴x + 4Sin²xCos²x = 24Sin²x(Sin²x + Cos²x) = 24Sin²x = 22Sin²x = 1Sin²x = 1/2Sinx = √2/2                                                 Sinx = - √2/2x = (-1)ⁿarcSin√2/2 + πn, n ∈ z                  x = (-1)ⁿarcSin(-√2/2) + πn, n ∈ zx = (-1)ⁿπ/4 + πn, n ∈ z                              x = (-1)ⁿ(-π/4) + πn, n ∈ z                                                                   x = (-1)ⁿ⁺¹π/4 + πn, n ∈ zКорни из заданного промежутка : - 7π/4 , ...