ответ: - 5π/3, - 2π, -3π.
Пошаговое объяснение:
√3sinx + 2sin(2x + π/6) = √3sin2x + 1
√3sinx + 2sin2x · cos(π/6) + 2cos2x · sin(π/6) = √3sin2x + 1
√3sinx + 2sin2x · √3/2 + 2 · cos2x · 1/2 = √3sin2x + 1
√3sinx + √3sin2x + cos2x = √3sin2x + 1
√3sinx + cos2x = 1
√3sinx + 1 - 2sin²x = 1
2sin²x - √3sinx = 0
sinx (2sinx - √3) = 0
1) sinx = 0
x = πn, n∈Z
2) sinx = √3/2
x = π/3 + 2πk, k∈Z x = 2π/3 + 2πm, m∈Z
x ∈ [- 3π; - 3π/2]:
1)
-3π ≤ πn ≤ -3π/2
-3 ≤ n ≤ -1,5
n ∈ Z, ⇒ n = - 3 x = - 3π
n = - 2 x = - 2π
2)
- 3π ≤ π/3 + 2πk ≤ - 3π/2
- 10π/3 ≤ 2πk ≤ - 11π/6
- 5/3 ≤ k ≤ - 11/12
k ∈ Z, ⇒ k = - 1 x = - 5π/3
- 3π ≤ 2π/3 + 2πm ≤ - 3π/2
- 11π/3 ≤ 2πm ≤ - 13π/6
- 11/6 ≤ m ≤ - 13/12
m ∈ Z, нет целых значений m на промежутке.
(16 - 4)/2 = 6