Question

а) sinx + 2sinx(2x + π/6)=√3sin2x + 2
б) корни на отрезке [-7π/2; -2π]

Answer 1

Ответ:

х₁ = πn;   x₂ = π/6 + 2πk;   x₃ = 5π/6 + 2πm;   n, k, m ∈ Z

Корни - 19π/6; -3π;  -2π;

Объяснение:

sinx + 2sin(2x + π/6) = √3sin2x + 1

sinx + 2(sin2x · cos π/6 + cos2x · sin π/6) = √3sin2x + 1

sinx + 2(sin2x · 0.5√3 + cos2x ·0.5) = √3sin2x + 1

sinx + √3sin2x + cos2x  = √3sin2x + 1

sinx  + cos2x  =  1

sinx  + 1 - 2sin²x  =  1

2sin²x - sinx = 0

sinx (2sinx - 1) = 0

1) sin x = 0

x = πn

б) корни на интервале [-7π/2; -2π]

--7π/2 ≤ πn ≤ -2π

-3.5 ≤ n ≤ -2

n = -3  и n = -2

корни -3π и -2π

2) 2sinx - 1 = 0

sinx = 1/2

x₂ = π/6 + 2πk

x₃ = 5π/6 + 2πm

б) корни на интервале [-7π/2; -2π]

-7π/2 ≤ π/6 + 2πk ≤ -2π

-7/2 - 1/6 ≤  2k ≤ -2 - 1/6

-22/6 ≤ 2k ≤ -13/6

-11/6 ≤ k ≤ -13/12

$-1\frac{5}{6}$ ≤ k ≤ $-1\frac{1}{12}$

Корней нет

-7π/2 ≤ 5π/6 + 2πm ≤ -2π

-7/2 - 5/6 ≤  2m ≤ -2 - 5/6

-26/6 ≤ 2m ≤ -17/6    

-13/6 ≤ m ≤ -17/12

$-2\frac{1}{6}$ ≤ m ≤ $-1\frac{5}{12}$

m = -2

Корень

5π/6 - 4π = - 19π/6