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ashixminas
@ashixminas
July 2022
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Помоги решить тригонометрическое уравнение!!!!!
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Область определения: sinx ≠0
x ≠ πn
cos2x(4cosx - 1/sinx) = 0
1) cos2x = 0 2) (4cosx · sinx - 1)/sinx = 0
2x = π/2 + πk 4cosx · sinx = 1
x = π/4 + πk/2 2sin2x = 1
sin2x = 1/2
2x = π/6 + 2πt или 2x = 5π/6 + 2πm
x = π/12 + πt x = 5π/12 + πm
[π ; 3π/2]
π ≤ π/4 + πk/2 ≤ 3π/2 π ≤ π/12 + πt ≤ 3π/2 π ≤ 5π/12 + πm ≤ 3π/2
4 ≤ 1 + 2k ≤ 6 12 ≤ 1 + 12t ≤ 18 12 ≤ 5 + 12m ≤ 18
1,5 ≤ k ≤ 2,5 11/12 ≤ t ≤ 17/12 7/12 ≤ m ≤ 13/12
k = 2 t = 1 m = 1
x = 5π/4 x = π/12 + π = 13π/12 x = 5π/12 + π = 17π/12
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Answers & Comments
Verified answer
Область определения: sinx ≠0x ≠ πn
cos2x(4cosx - 1/sinx) = 0
1) cos2x = 0 2) (4cosx · sinx - 1)/sinx = 0
2x = π/2 + πk 4cosx · sinx = 1
x = π/4 + πk/2 2sin2x = 1
sin2x = 1/2
2x = π/6 + 2πt или 2x = 5π/6 + 2πm
x = π/12 + πt x = 5π/12 + πm
[π ; 3π/2]
π ≤ π/4 + πk/2 ≤ 3π/2 π ≤ π/12 + πt ≤ 3π/2 π ≤ 5π/12 + πm ≤ 3π/2
4 ≤ 1 + 2k ≤ 6 12 ≤ 1 + 12t ≤ 18 12 ≤ 5 + 12m ≤ 18
1,5 ≤ k ≤ 2,5 11/12 ≤ t ≤ 17/12 7/12 ≤ m ≤ 13/12
k = 2 t = 1 m = 1
x = 5π/4 x = π/12 + π = 13π/12 x = 5π/12 + π = 17π/12