Предмет: Алгебра, автор: marina25658

2sin в квадрате x-4cos в квадрате x=7sinx*cosx
помогите пожалуйста,ОЧЕНЬ СРОЧНО

Ответы

Автор ответа: sharadi
0
2Sin² x-4Cos² x=7Sinx*Cosx | : Cos²x≠0
2tg²x -4 - 7tgx = 0
tgx = t
2t² -7t -4 = 0
D = 49 -4*2*(-4) = 81
t₁ = 4,                                 t₂ = -1/2
tgx = 4                                  tgx = -arctg1/2 + πn , n ∈Z
x = arctg4 + πk , k ∈Z
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