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clays4
@clays4
August 2022
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Всего лишь построить график !20 БАЛОВ
Постройте график функции у=1/2 sin2х
Постройте график функции у= arcsin(x-1)-1
Найдите корни уравнения sin(4х/3+п/6)=-1/2 принадлежащие промежутку [-2п;2п)
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dnepr1
Verified answer
1)
График функции у=(1/2) sin2х в приложении.
2) Г
рафик функции у= arcsin(x-1)-1 в приложении.
3) Дано уравнение sin((4х/3)+(π/6))=-1/2 и промежуток [-2π;2π
).
(4х/3)+(π/6) = 2
πk - (π/6),
(4х/3) = 2
πk -(2π/6),
(8х/6) = (12
πk/6) - (2π/6),
8х = 12πk - 2π,
x = 12πk/8 - (2π/8),
x = (3πk/2) - (π/4), k ∈ Z.
(4х/3)+(π/6) = 2
πk - (5π/6),
(4х/3) = 2
πk -(6π/6),
(8х/6) = (12
πk/6) - (6π/6),
8х = 12πk - 6π,
x = 12πk/8 - (6π/8),
x = (3πk/2) - (3π/4), k
∈ Z.
В заданном промежутке 5 корней:
х =-5,49779,
х = -2,35619,
х = -0,785398,
х = 2,35619,
х = 3,92699.
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Verified answer
1) График функции у=(1/2) sin2х в приложении.2) График функции у= arcsin(x-1)-1 в приложении.
3) Дано уравнение sin((4х/3)+(π/6))=-1/2 и промежуток [-2π;2π).
(4х/3)+(π/6) = 2πk - (π/6),
(4х/3) = 2πk -(2π/6),
(8х/6) = (12πk/6) - (2π/6),
8х = 12πk - 2π,
x = 12πk/8 - (2π/8),
x = (3πk/2) - (π/4), k ∈ Z.
(4х/3)+(π/6) = 2πk - (5π/6),
(4х/3) = 2πk -(6π/6),
(8х/6) = (12πk/6) - (6π/6),
8х = 12πk - 6π,
x = 12πk/8 - (6π/8),
x = (3πk/2) - (3π/4), k ∈ Z.
В заданном промежутке 5 корней:
х =-5,49779,
х = -2,35619,
х = -0,785398,
х = 2,35619,
х = 3,92699.