x1 = -3Π/2; x2 = -Π/2; x3 = -5Π/3
Пошаговое объяснение:
2sin(x+П/3) - √3*cos(2x) = sin x + √3
2(sin x*cos(Π/3) + cos x*sin(Π/3)) - √3*(2cos^2 x - 1) = sin x + √3
2sin x*1/2 + 2cos x*√3/2 - 2√3*cos^2 x + √3 = sin x + √3
2sin x*1/2 = sin x и √3 сокращаются.
√3*cos x - 2√3*cos^2 x = 0
-√3*cos x*(2cos x - 1) = 0
1) cos x = 0; x1 = Π/2 + Π*k, k € Z
2) 2cos x - 1 = 0
cos x = 1/2; x2 = Π/3 + 2Π*n; x3 = -Π/3 + 2Π*n, n € Z
На промежутке [-2П; - П/2] будут корни:
А) -2Π ≤ П/2 + П*k ≤ -Π/2
-2 ≤ 1/2 + k ≤ -1/2
- 2 1/2 ≤ k ≤ -1/2 - 1/2
-2,5 ≤ k ≤ -1
k € Z, поэтому k = -2; -1
x1 = П/2 - 2Π = -3Π/2; x2 = Π/2 - Π = -Π/2
Б) -2Π ≤ Π/3 + 2Π*n ≤ -Π/2
-2Π - Π/3 ≤ 2Π*n ≤ - Π/2 - Π/3
- 2 1/3 ≤ 2n ≤ - 5/6
- 1 1/6 ≤ n ≤ -5/12
n € Z, поэтому n = -1
x3 = Π/3 - 2Π = - 5Π/3
В) -2Π ≤ -Π/3 + 2Π*n ≤ -Π/2
-2Π + Π/3 ≤ 2Π*n ≤ -Π/2 + Π/3
-2 + 1/3 ≤ 2n ≤ -1/6
-5/6 ≤ n ≤ -1/12
На этом промежутке корней нет.
