Задача 61004 ...
Условие
Решение
cos(x/2)·sin(3x/2)+sin(x/2)·cos(3x/2)=4·sin^2 (π+x)cos^2 (π–x)
cos(x/2)·sin(3x/2)+sin(x/2)·cos(3x/2)=sin((3x/2)+(x/2))=sin2x
sin (π+x)=-sinx
sin^2 (π+x)=(-sinx)^2=sin^2x
cos (π–x)=-cosx
cos^2 (π–x)=(-cosx)^2=cos^2x
4·sin^2 (π+x)cos^2 (π–x)=4sin^2x *cos^2x=(2*sinx*cosx)^2=sin^22x
sin2x=sin^22x ⇒
sin2x =0 или sin2x=1
2x =πk, k ∈ Z или 2x=(π/2)+2πn, n ∈ Z
x =(π/2)*k, k ∈ Z или x=(π/4)+πn, n ∈ Z
Указанному промежутку [π; 3π] принадлежат корни:
(π/2)*2=π
(π/2)*3=3π/2
(π/2)*4=2π
(π/2)*5=5π/2
(π/2)*6=3π
(π/4)+π=5π/4
(π/4)+2π=9π/4