Найти корни уравнений 1. Найти наименьший положительный
cos (pi(x + 2))/6 = (sqrt(2))/2
sin x + sin 13x = 0
2. Наибольший отрицательный
cos(pi(5x+4)/6 = -(1/2)
2sin²((pi*x)/12) + 3 cos((pi*x)/12)=0
2 sin² x + 3 sin x cos x + cos²x = 0

Предыдущий
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вопрос

cos(pi(x + 2))/6 = sqrt(2)/2
cos(pi(x + 2)) = 3sqrt(2)/2
pi(x + 2) = arccos(3sqrt(2)/2) + 2kpi or pi(x + 2) = -arccos(3sqrt(2)/2) + 2kpi
x = (arccos(3sqrt(2)/2) + 2kpi - 2) / pi or x = (-arccos(3sqrt(2)/2) + 2kpi - 2) / pi

sin x + sin 13x = 0
sin x + sin 13x = 2sin(7x)cos(6x) = 0
sin(7x) = 0 or cos(6x) = 0
x = kpi, k in Z or x = (2n + 1)pi/4, n in Z

cos(pi(5x+4)/6) = -1/2
pi(5x+4)/6 = +/- 2pi/3 + 2kpi or pi(5x+4)/6 = pi + 2kpi
x = (4 +/- 6)/5 or x = (18 - 4)/5
x = 2/5 or x = 14/5

2sin²((pix)/12) + 3 cos((pix)/12) = 0
2sin²((pix)/12) = -3cos((pix)/12)
tan((pix)/12) = -3/2
(pix)/12 = arctan(-3/2) + kpi or (pi*x)/12 = arctan(-3/2) + pi/2 + kpi
x = 12(arctan(-3/2) + kpi)/pi or x = 12(arctan(-3/2) + pi/2 + kpi) / pi

2 sin² x + 3 sin x cos x + cos² x = 0
sin x (2sin x + 3cos x) + cos² x = 0
sin x (2sin x + 3cos x) + 1 - sin² x = 0
sin x (2sin x + 3cos x) = sin² x
2sin² x + 3sin x cos x = sin² x
sin x (2sin x + 3cos x - sin x) = 0
sin x = 0 or 2sin x + 3cos x - sin x = 0
x = kpi, k in Z or 3cosa - sin = 0
x = arctan(3) + kpi or x = pi - arctan(1/3) + kpi