Задача 32434 (sin2x-sqrt(3)cos2x)^2 - 5 = cos(Pi/6 -...

vk177899689

2018-12-25 21:46:24

(sin2x-sqrt(3)cos2x)^2 - 5 = cos(Pi/6 - 2x)

sova

2018-12-25 22:58:39

cos((π/6)-2x)= cos(π/6)*cos2x - sin(π/6)*sin2x =

=sqrt(3)/2*cos2x-(1/2)si2x=(1/2)*(sqrt(3)cos2x-sin2x);

Замена
sin2x-sqrt(3)cos2x=t

t^2+(1/2)*t-5=0
2t^2+t-10=0
D=1-4*2*(-10)=81
t_(1)=(-1-9)/4=-5/2 или t_(2)=(-1+9)/2=2

sin2x-sqrt(3)cos2x=- 5/2 или sin2x-sqrt(3)cos2x=2

sin2x-sqrt(3)cos2x= 5/2 - уравнение не имеет корней

(1/2)sin2x-sqrt(3)/2cos2x=-5/4


sin(2x-(π/6))=-5/4 не имеет корней, так как |sin(2x-(π/6)| ≤ 1


sin2x-sqrt(3)cos2x=2

(1/2)sin2x-(sqrt(3)/2)cos2x=1
sin(2x-(π/6))=1

2x-(π/6)=(π/2)+2πn, n ∈ Z
2x=(π/2)+(π/6)+2πn, n ∈ Z
2x=(2π/3)+2πn, n ∈ Z
[b]x=(π/3)+2πn, n ∈ Z[/b]