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MacTy
@MacTy
July 2022
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a) Решите уравнение -cos2x-1=3cos(7П/2 -x)
б)Укажите корни этого уравнения , принадлежащие отрезку [-5П/2 ; -П]
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flsh
Verified answer
-cos 2x - 1 = 3cos(7π/2 - x)
-cos 2x - 1 = 3cos(3π/2 - x)
-cos 2x - 1 = -3sin x
cos 2x + 1 = 3sin x
1 - 2sin^2 x + 1 = 3sin x
2sin^2 x + 3sin x - 2 = 0
Замена: sin x = t
2t^2 + 3t - 2 = 0
D = 9 + 16 = 25
t1 = (-3 - 5)/4 = -2
t2 = (-3 + 5)/4 = 1/2
sin x = -2 - нет решений
sin x = 1/2
x = (-1)^n * π/6 + πn, n € Z
Корни, принадлежащие указанному отрезку:
-11π/6, -7π/6.
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MacTy
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Answers & Comments
Verified answer
-cos 2x - 1 = 3cos(7π/2 - x)-cos 2x - 1 = 3cos(3π/2 - x)
-cos 2x - 1 = -3sin x
cos 2x + 1 = 3sin x
1 - 2sin^2 x + 1 = 3sin x
2sin^2 x + 3sin x - 2 = 0
Замена: sin x = t
2t^2 + 3t - 2 = 0
D = 9 + 16 = 25
t1 = (-3 - 5)/4 = -2
t2 = (-3 + 5)/4 = 1/2
sin x = -2 - нет решений
sin x = 1/2
x = (-1)^n * π/6 + πn, n € Z
Корни, принадлежащие указанному отрезку:
-11π/6, -7π/6.