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зина25
@зина25
August 2022
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(cos2x+cosp÷4(cos2x+4)=0и найдите его корни принадлежащие промежуткк [0;p]
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Answers & Comments
irka1804
Cosπ = -1
cos2x - 1/ (cos2x + 4) = 0, cos2x + 4 > 0, поэтому домножим на это
cos2x(cos2x + 4) - 1 = 0
cos^2(2x) + 4cos2x - 1 = 0
t = cos2x
t^2 +4t - 1 = 0
t = (-4 +-√(16 +4)) / 2 = (-4 +-2√5) / 2 = -2 +-√5
t = -2 -√5 не подходит, так как cos2x >= -1
t = -2 +√5
cos2x = -2 + √5
2x = +-arccos(√5 - 2) + 2πn, n ∈ Z
x = +-arccos(√5 - 2) / 2 + πn, n ∈ Z
0 <= +-arccos(√5 - 2) / 2 + πn <=
π
x =
arccos(√5 - 2) / 2
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Answers & Comments
cos2x - 1/ (cos2x + 4) = 0, cos2x + 4 > 0, поэтому домножим на это
cos2x(cos2x + 4) - 1 = 0
cos^2(2x) + 4cos2x - 1 = 0
t = cos2x
t^2 +4t - 1 = 0
t = (-4 +-√(16 +4)) / 2 = (-4 +-2√5) / 2 = -2 +-√5
t = -2 -√5 не подходит, так как cos2x >= -1
t = -2 +√5
cos2x = -2 + √5
2x = +-arccos(√5 - 2) + 2πn, n ∈ Z
x = +-arccos(√5 - 2) / 2 + πn, n ∈ Z
0 <= +-arccos(√5 - 2) / 2 + πn <= π
x = arccos(√5 - 2) / 2