Предмет: Алгебра,
автор: supermee
Решите уравнение:
1-2cos^2(4x-п/4)=0
Ответы
Автор ответа:
0
1 - 2cos²(4x - π/4) = 0
cos²(4x - π/4) = 1/2
cos(4x - π/4) = ±√2/2
1) cos(4x - π/4) = √2/2
4x - π/4 = ±π/4 + 2πn, n ∈ Z
4x = ±π/4 + π/4 + 2πn, n ∈ Z
x = ±π/16 + π/16 + πn/2, n ∈ Z
2)cos(4x - π/4) = -√2/2
4x - π/4 = ±3π/4 + 2πn, n ∈ Z
4x = ±3π/4 + π/4 + 2πn, n ∈ Z
x = ±3π/16+ π/16 + πn/2, n ∈ Z
cos²(4x - π/4) = 1/2
cos(4x - π/4) = ±√2/2
1) cos(4x - π/4) = √2/2
4x - π/4 = ±π/4 + 2πn, n ∈ Z
4x = ±π/4 + π/4 + 2πn, n ∈ Z
x = ±π/16 + π/16 + πn/2, n ∈ Z
2)cos(4x - π/4) = -√2/2
4x - π/4 = ±3π/4 + 2πn, n ∈ Z
4x = ±3π/4 + π/4 + 2πn, n ∈ Z
x = ±3π/16+ π/16 + πn/2, n ∈ Z
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