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elena611
@elena611
July 2022
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помогите пожалуйста решить cos4x-sin2x=0
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Verified answer
Cos4x-sin2x=0
Формула cos 2α = 1 - 2sin²α
1 - 2sin² 2x - sin 2x = 0
2sin²2x + sin 2x - 1 = 0 - квадратное уравнение с неизвестным sin(2x)
D = 1 + 4*2 = 9 = 3²
1) sin 2x = (-1 + 3)/4 = 1/2;
2x = π/6 + 2πn; или 2x = 5π/6 + 2πk;
x₁ = π/12 + πn; x₂ = 5π/12 + πk
2) sin 2x = (-1 - 3)/4 = -1
2x = -π/2 + 2πm;
x₃ = -π/4 + πm
Ответ: x₁ = π/12 + πn; x₂ = 5π/12 + πk; x₃ = -π/4 + πm;
n,k,m ∈ Z
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Answers & Comments
Verified answer
Cos4x-sin2x=0Формула cos 2α = 1 - 2sin²α
1 - 2sin² 2x - sin 2x = 0
2sin²2x + sin 2x - 1 = 0 - квадратное уравнение с неизвестным sin(2x)
D = 1 + 4*2 = 9 = 3²
1) sin 2x = (-1 + 3)/4 = 1/2;
2x = π/6 + 2πn; или 2x = 5π/6 + 2πk;
x₁ = π/12 + πn; x₂ = 5π/12 + πk
2) sin 2x = (-1 - 3)/4 = -1
2x = -π/2 + 2πm;
x₃ = -π/4 + πm
Ответ: x₁ = π/12 + πn; x₂ = 5π/12 + πk; x₃ = -π/4 + πm;
n,k,m ∈ Z