Вложение!!!!
Решите уравнение 2cos^3x+корень3сos^2x+2cosx+корень3=0
Указать корни уравнения на отрезке [-2п;-п/2]
Решите уравнение 2cos^3x+корень3сos^2x+2cosx+корень3=0
Указать корни уравнения на отрезке [-2п;-п/2]
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Verified answer
2*cosx*(cos²x + 1) + √3*(cos²x + 1) = 0(cos²x + 1)*(2*cosx + √3) = 0
1) cos²x + 1 = 0
cos²x = -1
нет решений
2) 2*cosx + √3 = 0
cosx = -√3/2
x = +- 5 π/6 + 2πk
-2π ≤ -5π/6 + 2πk ≤ -π/2
-12 ≤- 5+12k ≤-3
-7 ≤ 12k ≤ 2
-7/12 ≤ k ≤ 1/6
k=0 x=-5π/6
-2π ≤ 5π/6 + 2πk ≤ -π/2
-12 ≤ 5 +12k ≤ -3
-17 ≤ 12k ≤ -8
-17/12 ≤ k ≤ -2/3
k=-1 x=5π/6-2π=-7π/6