3cos^2x - 5cos x = 2
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Ответил kopatel228
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3Cos^2x - 5Cosx - 2 = 0
t = Cosx ∈ [-1;1]
3t^2 - 5t - 2 = 0
D = 25 - 4 * 3 * (-2) = 49
t1 = 5 + 7 / 6 = 2 ∉ [-1;1]
t2 = 5 - 7 / 6 = -1/3
Cosx = -1/3
x = arccos(-1/3) + 2πn, n∈Z
t = Cosx ∈ [-1;1]
3t^2 - 5t - 2 = 0
D = 25 - 4 * 3 * (-2) = 49
t1 = 5 + 7 / 6 = 2 ∉ [-1;1]
t2 = 5 - 7 / 6 = -1/3
Cosx = -1/3
x = arccos(-1/3) + 2πn, n∈Z
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