2sin(x+π/3)-√3cos2x=sin x + √3
Корни на отрезке [-2π;-π/2]
С решением

Ответ:

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.