Решите уравнение
cos^2x-sin^2x=-1/2
тут формула: косинус двойного угла
cos2x = -1/2
2x = ±(π-arccos(1/2)) + 2πn, n ∈ z
2x = ±2π/3 + 2πn, n ∈ z | /2
x = ±π/3 + πn, n ∈ z
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тут формула: косинус двойного угла
cos2x = -1/2
2x = ±(π-arccos(1/2)) + 2πn, n ∈ z
2x = ±2π/3 + 2πn, n ∈ z | /2
x = ±π/3 + πn, n ∈ z