Автор - shugylatazhikul

4cos^2x/2+0,5sinx+3sin^2x/2=3

Ответ

Автор - zefirnuu

Ответ:

x = π + 2πn, n ∈ Z; -π/2 + 2πk, k ∈ Z.

Объяснение:

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

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