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kristina1903
@kristina1903
July 2022
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Решите уравнение cos6x-cos3x=0. Укажите корни, принадлежащие отрезку [0; пи].
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kalbim
Verified answer
Cos(6x) - cos(3x) = -2*sin(9x/2)*sin(3x/2) = 0
sin(9x/2) = 0, 9x/2=πk, x=2πk/9
sin(3x/2) = 0, 3x/2=πk, x=2πk/3
Найдем, при каких к корни будут принадлежать указанному промежутку:
0≤2πk/9≤π, 0≤k≤4.5 - т.е. k=0, 1, 2, 3, 4
0≤2πk/3≤π, 0≤k≤1.5 - т.е. k=0, 1
x∈[0;π]
k=0,
x=0
k=1,
x=2π/9, x=2π/3
k=2,
x=4π/9
, x=4π/3
k=3,
x=6π/9 = 2π/3
, x=6π/3 = 2π
k=4,
x=8π/9
Ответ:
0, 4π/9, 2π/3, 8π/9
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Answers & Comments
Verified answer
Cos(6x) - cos(3x) = -2*sin(9x/2)*sin(3x/2) = 0sin(9x/2) = 0, 9x/2=πk, x=2πk/9
sin(3x/2) = 0, 3x/2=πk, x=2πk/3
Найдем, при каких к корни будут принадлежать указанному промежутку:
0≤2πk/9≤π, 0≤k≤4.5 - т.е. k=0, 1, 2, 3, 4
0≤2πk/3≤π, 0≤k≤1.5 - т.е. k=0, 1
x∈[0;π]
k=0, x=0
k=1, x=2π/9, x=2π/3
k=2, x=4π/9, x=4π/3
k=3, x=6π/9 = 2π/3, x=6π/3 = 2π
k=4, x=8π/9
Ответ: 0, 4π/9, 2π/3, 8π/9