ПОМОГІТЕ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!11. Created by Kristina7942401. matematika-ru

ПОМОГІТЕ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!11...

July 2022 1 20 Report

ПОМОГІТЕ!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!11

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sedinalana 3
(1+сos4x)³/8+(1-cos4x)³/8=13/16
2(1+3cos4x+3cos²4x+cos³4x+1-3cos4x+3cos²4x-cos³4x)=13
2(2+6cos²4x)=13
4+12cos²4x=13
cos²4x=3/4
cos4x=-√3/2 U cos4x=√3/2
4x=-5π/6+2πk,k∈z ⇒x=-5π/24+πk/2,k∈z при k=0 x=-5π/24∈[-π/4;π/4]
4x=-π/6+2πk,k∈z ⇒x=-π/24+πk/2,k∈z при k=0 x=-π/24∈[-π/4;π/4]
4x=π/6+2πk,k∈z ⇒x=π/24+πk/2,k∈z при k=0 x=π/24∈[-π/4;π/4]
4x=5π/6+2πk,k∈z ⇒x=5π/24+πk/2,k∈z при k=0 x=5π/24∈[-π/4;π/4]
Ответ 4 корня
4
cosx=√sin²x/2
cosx=|sin x/2|, x∈[-π/2+2πk,π/2+2πk,k∈z
1)x∈[-π/2+2πk;2πk,k∈z]
cosx=-sinx/2
2sin²x/2-1-sin x/2=0
sin x/2=t
2t²-t-1=0
D=1+8=9
t1=(1-3)/4=-1/2⇒sinx/2=-1/2
x/2=-7π/6+2πk⇒x=-7π/3+4πk,k∈zне удов усл
x/2=-π/6+2πk⇒x=-π/3+4πk,k∈z
2)x∈[2πk;π/2+2πk,k∈z]
cosx=sinx/2
2sin²x/2-1+sin x/2=0
sin x/2=t
2t²+t-1=0
D=1+8=9
t1=(-1-3)/4=-1⇒sinx/2=-1
x/2=-π/2+2πk.k∈z⇒x=-2π+4πk,k∈z не удов усл
t2=(-1+3)/4=1/2
sinx/2=π/6+2πk,k∈z⇒x=π/3+4πk,k∈z
sinx/2=5π/6+2πk,k∈z⇒x=5π/3+4πk,k∈z не удов усл
-----------------------------------
найдем корни на заданном промежутке
x=-π/3+4πk,k∈z при k=0 x=-π/3
x=-π/2+4πk,k∈z при k=0 x=-π/2
x=π/3+4πk,k∈z при k=0 x=π/3
-π/3-π/2+π/3=-π/2=-90гр

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