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anutaerm
@anutaerm
July 2022
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48 баллов алгебра 10 класс помогите прошу
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Cos(x)*(1 - 2(sin(2x))^2) = 0
1) cos(x) = 0 ---в указанном отрезке корень один: х = 90°
решение: х = (pi/2) + pi*k, k∈Z
2) sin(2x) = +-1/√2 ---в указанном отрезке один корень: х = 3*180/8 = 67.5°
решение: 2х = +-(pi/4) + 2pi*k, k∈Z ---> x = +-(pi/8) + pi*k, k∈Z
2х = +-(3pi/4) + 2pi*k, k∈Z ---> x = +-(3pi/8) + pi*k, k∈Z
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Answers & Comments
Verified answer
Cos(x)*(1 - 2(sin(2x))^2) = 01) cos(x) = 0 ---в указанном отрезке корень один: х = 90°
решение: х = (pi/2) + pi*k, k∈Z
2) sin(2x) = +-1/√2 ---в указанном отрезке один корень: х = 3*180/8 = 67.5°
решение: 2х = +-(pi/4) + 2pi*k, k∈Z ---> x = +-(pi/8) + pi*k, k∈Z
2х = +-(3pi/4) + 2pi*k, k∈Z ---> x = +-(3pi/8) + pi*k, k∈Z