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oliviakir
@oliviakir
August 2022
1
24
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КТО - НИБУДЬ, ПОЖАЛУЙСТА!!!!
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2cos^2 x + sin x - 1 = 0
cos x >= 0
x € [300°; 700°]
2 - 2sin^2 x + sin x - 1 = 0
x € [-90° + 360°*n; 90° + 360°*n], n € Z
x € [300°; 700°]
2sin^2 x - sin x - 1 = 0
x € [300°; 450] U [630°; 700°]
2sin^2 x - sin x - 1 = 0
sin x = t
2t^2 - t - 1 = 0
D = 9
t1 = (1 - 3)/4 = -1/2
t2 = (1 + 3)/4 = 1
sin x = -1/2 => x = (-1)^(n + 1) * 30° + 180°*n, n € Z
sin x = 1 => x = 360°*n, n € Z
x = (-1)^(n + 1) * 30° + 180°*n, n € Z
x = 360°*n, n € Z
x € [300°; 450] U [630°; 700°]
x = 330°, 360°, 690°
Ответ: 1380
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Answers & Comments
cos x >= 0
x € [300°; 700°]
2 - 2sin^2 x + sin x - 1 = 0
x € [-90° + 360°*n; 90° + 360°*n], n € Z
x € [300°; 700°]
2sin^2 x - sin x - 1 = 0
x € [300°; 450] U [630°; 700°]
2sin^2 x - sin x - 1 = 0
sin x = t
2t^2 - t - 1 = 0
D = 9
t1 = (1 - 3)/4 = -1/2
t2 = (1 + 3)/4 = 1
sin x = -1/2 => x = (-1)^(n + 1) * 30° + 180°*n, n € Z
sin x = 1 => x = 360°*n, n € Z
x = (-1)^(n + 1) * 30° + 180°*n, n € Z
x = 360°*n, n € Z
x € [300°; 450] U [630°; 700°]
x = 330°, 360°, 690°
Ответ: 1380