A)4cos^2x/2+0,5sinx+3sin^2x/2=3 b)sin^4x-cos^4x=0,5

nikitakyznecov703 nikitakyznecov703    2   13.09.2019 08:50    0

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marinandnatasha1 marinandnatasha1  07.10.2020 11:32
А) 4cos²(x/2) + 0,5sinx + 3sin²(x/2) = 3
4cos²(x/2) + 2·0,5sin(x/2)·cos(x/2) + 3sin²(x/2) = 3sin²(x/2) + 3cos²(x/2) 
cos²(x/2) + sin(x/2)cos(x/2) = 0
cos(x/2)[cos(x/2) + sin(x/2)] = 0
1) cos(x/2) = 0
x/2 = π/2 + πn, n ∈ Z
x = π + 2πn, n ∈ Z
2) cos(x/2) + sin(x/2) = 0
cos(x/2) = -sin(x/2)
tg(x/2) = -1
x/2 = -π/4 + πk, k ∈ Z
x = -π/2 + 2πk, k ∈ Z
ответ: x = π + 2πn, n ∈ Z; -π/2 + 2πk, k ∈ Z.

б) sin⁴x - cos⁴x = 0,5
(sin²x - cos²x)(sin²x + cos²x) = 0,5
sin²x - cos²x = 0,5
-cos2x = 1/2
cos2x = -1/2
2x = ±2π/3 + 2πn, n ∈ Z
x = ±π/3 + πn, n ∈ Z
ответ: x = ±π/3 + πn, n ∈ Z.
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