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sssergei
@sssergei
July 2022
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Решите уравнение
1) 1-cosx=2sin x/2
2) 2sin^2 x/2 +1/2sin2x =1
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kalbim
Verified answer
1) 1-cosx = 2sin(x/2)
1-cosx = 1-cos(2*x/2) = 1 - (1-2sin^2(x/2)) = 2sin^2(x/2)
2sin^2(x/2) = 2sin(x/2)
2sin^2(x/2) - 2sin(x/2) = 0
sin(x/2) * (sin(x/2) - 1) = 0
a) sin(x/2) = 0, x/2 = pi*k, x=2pi*k
b) sin(x/2) = 1, x/2 = pi + 2pi*k
Объединив решения, получается: x=pi*k
2) 2sin^2(x/2) = 1 - cosx
1-cosx + cosx*sinx = 1
cosx(sinx-1)=0
a) cosx=0, x=pi/2 + pi*k
b) sinx=1, x=pi/2+pi*k
Общее решение: x=pi/2 + pi*k
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Answers & Comments
Verified answer
1) 1-cosx = 2sin(x/2)1-cosx = 1-cos(2*x/2) = 1 - (1-2sin^2(x/2)) = 2sin^2(x/2)
2sin^2(x/2) = 2sin(x/2)
2sin^2(x/2) - 2sin(x/2) = 0
sin(x/2) * (sin(x/2) - 1) = 0
a) sin(x/2) = 0, x/2 = pi*k, x=2pi*k
b) sin(x/2) = 1, x/2 = pi + 2pi*k
Объединив решения, получается: x=pi*k
2) 2sin^2(x/2) = 1 - cosx
1-cosx + cosx*sinx = 1
cosx(sinx-1)=0
a) cosx=0, x=pi/2 + pi*k
b) sinx=1, x=pi/2+pi*k
Общее решение: x=pi/2 + pi*k