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Вася20006
@Вася20006
November 2021
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нужен только 18 и 24, заранее благодарен
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sedinalana
Verified answer
18
sin12x=1/2*sinx-√3/2*cosx
sin12x=sin(x-π/3)
sin12x-sin(x-π/3)=0
2sin(11x/2+π/6)cos(13x/2-π/6)=0
sin(11x+π/6)=0⇒11x/2+π/6=πn⇒11x/2=-π/6+πn⇒x=-π/33+2πn/11,n∈z
cos(13x/2-π/6)=0⇒13x/2-π/6=π/2+πk⇒13x/2=2π/3+πk⇒
x=4π/39+2πk/13,k∈z
24
2cos²(x/2)-1+4cos(x/2)-5=0
cos(x/2)=a
2a²+4a-6=0
a²+2a-3=0
a1+a2=-2 U a1*a2=-3
a1=-3⇒cos(x/2)=-3<-1 нет решения
a2=1⇒cos(x/2)=1⇒x/2=2πn⇒x=4πn,n∈z
1 votes
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Вася20006
Покорнейше благодарен. Вы только что спасли меня от двойке по математике за год.
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Answers & Comments
Verified answer
18sin12x=1/2*sinx-√3/2*cosx
sin12x=sin(x-π/3)
sin12x-sin(x-π/3)=0
2sin(11x/2+π/6)cos(13x/2-π/6)=0
sin(11x+π/6)=0⇒11x/2+π/6=πn⇒11x/2=-π/6+πn⇒x=-π/33+2πn/11,n∈z
cos(13x/2-π/6)=0⇒13x/2-π/6=π/2+πk⇒13x/2=2π/3+πk⇒
x=4π/39+2πk/13,k∈z
24
2cos²(x/2)-1+4cos(x/2)-5=0
cos(x/2)=a
2a²+4a-6=0
a²+2a-3=0
a1+a2=-2 U a1*a2=-3
a1=-3⇒cos(x/2)=-3<-1 нет решения
a2=1⇒cos(x/2)=1⇒x/2=2πn⇒x=4πn,n∈z