b)какие корни принадлежат отрезку [-5π;-7π/2]
Формула:
[b]sin α *cos β =(1/2)sin( α + β ) +(1/2)sin( α - β )
[/b]
cos(x+(π/6))*sinx= (1/2)sin(x+x+(π/6)) + (1/2)sin(x-x-(π/6)
cos(x+(π/6))*sinx=(1/2)sin(2x+(π/6))+(1/2)sin(-π/6)
cos(x+(π/6))*sinx=(1/2)sin(2x+(π/6))-(1/4)
Уравнение:
sin(2x+(π/6))=cosx+(1/2)sin(2x+(π/6))-(1/4)
[b](1/2)*sin(2x+(π/6))=cosx-(1/4)[/b]
Формула:
[b]sin( α + β )=sin α cos β +cos α sin β [/b]
(1/2)*sin2x*cos(π/6)+(1/2)*cos2xsin(π/6)=cosx-(1/4)
(1/2)*sin2x*(sqrt(3)/2)+(1/2)*cos2x*(1/2)=cosx-(1/4)
Умножаем на 4:
sqrt(3)sin2x + cos2x=4cosx-1;
2sqrt(3)sinx*cosx+2cos^2x-1=4cosx-1;
2sqrt(3)sinx*cosx+2cos^2x-4cosx=0
2cosx*(sqrt(3)sinx+cosx-2)=0
cosx=0 или sqrt(3)sinx+cosx-2=0
cosx=0 ⇒ [b] x=(π/2)+πn, n ∈ Z[/b]
или
sqrt(3)sinx+cosx-2=0
sqrt(3)sinx+cosx=2
уравнение вида
asinx+bcosx=c
Решаем [b] либо методом введения вспомогательного угла[/b]
sqrt(3)/2*sinx+(1/2) cosx=1
cos(x-(π/3))=1
x-(π/3)=2πm, m ∈ Z
[b]х = (π/3)+2πm, m ∈ Z[/b]
[b]либо как однородное второго порядка[/b] с аргументом (x/2)
2sqrt(3)sin(x/2)*cos(x/2) +cos^2(x/2)-sin^2(x/2)=2*(cos^2(x/2)+sin^2(x/2))
2sqrt(3)sin(x/2)*cos(x/2) -cos^2(x/2)-3sin^2(x/2)=0
3tg^2(x/2)-2sqrt(3)tg(x/2)+1=0
D=12-12=0
tg(x/2)=sqrt(3)/3
(x/2)=(π/6)+πk, k ∈ Z
[b]x=(π/3)+2πk, k ∈ Z[/b]
О т в е т.
а) (π/2)+πn, n ∈ Z; (π/3)+2πk, k ∈ Z
б)
x=(π/2)-5π=-9π/2
x=(π/2)-4π=-7π/2
x=(π/3)-4π=-11π/3.
[b]-4π[/b]=-24π/6 < [b]-11π/3[/b]=-22π/6 [b] <[/b] [b] -7π/2[/b]=-21π/6