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ShKs
@ShKs
August 2022
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Решите уравнение
(Корень из 3*sin2x+cos2x)/2=1
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nKrynka
(√3/2)*sin2x + (1/2)*cos2x = 1
sin(π/3)sin2x + cos(π/3)cos2x = 1
cos(π/3 - 2x) = 1
cos(2x - π/3) = 1
2x - π/2 = 2πk, k∈Z
2x = π/2 + 2πk, k∈Z
x = π/4 + πk, k∈Z
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Answers & Comments
sin(π/3)sin2x + cos(π/3)cos2x = 1
cos(π/3 - 2x) = 1
cos(2x - π/3) = 1
2x - π/2 = 2πk, k∈Z
2x = π/2 + 2πk, k∈Z
x = π/4 + πk, k∈Z