13. а) Решите уравнение 2cos(2x -π/3) -sinx =√3sin2x .
13. б) Найдите все корни данного уравнения принадлежащие отрезку [ -5π ; -7π/2] .
ответ: а) - π/2 +2πk , k∈ ℤ , π/6 + 2πn , 5π/6 + 2πn , n ∈ℤ
б) - 4,5π ; - 23π/6
Пошаговое объяснение: 2cos(2x -π/3) - sinx =√3sin2x⇔
2*( cos2x*cos(π/3) +sin2x*sin(2π/3) ) - sinx =√3sin2x ⇔
2*( cos2x*1/2 +sin2x*√3/2 ) - sinx =√3sin2x ⇔
cos2x + √3sin2x - sinx = √3sin2x ⇔ 1 -2sin²x -sinx =0 ⇔
2sin²x + sinx - 1 =0 ⇒ sinx = - 1 ; sinx = 1/2
x = -π/2 +2πк , k∈ ℤ ; x = (-1)ⁿπ/6 + πn , n ∈ℤ иначе
x = -π/2 +2πk , k∈ ℤ ; x = π/6 + 2πn , x = 5π/6+ 2πn , n ∈ℤ
- - - - - - -
б) x ∈ [ -5π ; -7π/2] .
- 5π ≤ - π/2 +2πk ≤ -7π/2 ⇔ - 4,5π ≤ 2πk ≤ -3π ⇔ -2,25 ≤ k ≤ -1,5 ;
k = -2 ⇒ x = - 4,5π
- - -
x = π/6 + 2πn
n = - 2 ⇒ x = π/6+ 2π*(-2) = -4π+π/6 = -23π/6
n = -3 ⇒ x = π/6+ 2π*(-3) = -6π+π/6 = -35π/6 < - 5π
* * * - 5π ≤ π/6 +2πn ≤ -7π/2⇔-5π -π/6≤ 2πn ≤ -7π/2 -π/6 ⇔-31/12 ≤ n ≤ -22/12 n = -2 * * * - - - - - -
x = 5π/6 + 2πn ; в отрезке [ -5π ; - 7π/2 ] не содержит решения
действительно , выбираем :
n = - 3 ⇒ x =5π/6 + 2π*(-3) = -6π +5π/6 = -31π/6 < -5π = - 30π/6 ,
n= - 2 ⇒ x = - 19π/6 > - 7π/2 = - 21π/6
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
* * * - 5π ≤ 5π/6 +2πn ≤ -7π/2⇔ -5π -5π/6 ≤ 2πn≤ -7π/2-5π/6⇔
-35π/6≤ 2πn ≤ -26π/6 ⇔ - 35/12 ≤ n ≤ - 26/12 нет целое число * * *
x₁=π; x₁=2π; x₁=3π; x₂=5/6π; x₂=17π/6;
Пошаговое объяснение:
(2sin²x-sinx)/(2cosx-√3)=0; [3/2π; 3π];
2sin²x-sinx=0;
2cosx-√3≠0;
2sin²x-sinx=0; sinx=z;
2z²-z=0;
z(2z-1)=0;
z₁=0; z₂=1/2;
sinx=0; sinx=1/2;
x₁=arcsin0; x₂= (-1)ⁿarcsin1/2+πn;
x₁=0+πn; x₂= (-1)ⁿ*π/6+πn n∈Z
2cosx-√3≠0;
cosx≠√3/2;
x≠±arccos √3/2+2πn;
x≠±π/6+2πn n∈Z
x₁=0+πn;
x₂= (-1)ⁿ*π/6+πn
x≠±π/6+2πn n∈Z
[3/2π; 3π]
n=0; x₁=0; x₂=π/6; x≠π/6 ∉ [3/2π; 3π]
n=1; x₁=π; x₂=-π/6+π=5/6π; x≠±π/6+2π=13/6π;
n=2; x₁=2π; x₂=π/6+2π=13π/6; x≠±π/6+4π=25π/6;
n=3; x₁=3π; x₂=-π/6+3π=17π/6;
n=4; x₁=4π; x₂=π/6+4π=25π/6 ∉ [3/2π; 3π]
x₁=π; x₁=2π; x₁=3π; x₂=5/6π; x₂=17π/6;
900-12%
Х/900=100/12(перемножаем крест на крест)
12Х=90000
Х=7500