ПОМОГИТЕ ПОЖАЛУЙСТА!!! ОЧЕНЬ НУЖНО. Created by katyadinner. algebra-ru

Решение задания во вложении.

ПОМОГИТЕ ПОЖАЛУЙСТА!!! ОЧЕНЬ НУЖНО...

sedinalana 1
y=2sinx-√3x+√3/6π+7
y`=2cosx-√3=0
cosx=√3/2
x=π/6+2πk,k∈z
при k=0 x=π/6∈[0;π/2]
y(0)=2sin0-√3*0+√3/6π+7≈7,91
y(π/6)=2sinπ/6-√3*π/6+√3π/6+7=2*1/2+7=8 наиб
y(π/2)=2sinπ/2-√3π/2+√3π/6+7=2-√3π/3+7≈7,2
2
сos(3π/2+2x)=cosx
sin2x-cos=0
2sinxcosx-cosx=0
cosx*(2sinx-1)=0
cosx=0πx=π/2+πk,k∈z
sinx=1/2⇒x=π/6+2πk,k∈z U x=5π/6+2πk,k∈z

-5π/2≤π/2+πk≤4π
-5≤1+2k≤8
-6≤2k≤7
-3≤k≤3,5
k=-3⇒x=π/2-3π=-5π/2
k=-2⇒x=π/2-2π=-3π/2
k=-1⇒x=π/2-π=-π/2
k=0⇒x=π/2
k=1⇒x=π/2+π=3π/2
k=2⇒x=π/2+2π=5π/2
k=3⇒x=π/2+3π=5π/2

-5π/2≤π/6+2πk≤4π
-15≤1+12k≤24
-16≤12k≤23
-16/12≤k≤23/12
k=-1⇒x=π/6-2π=-11π/6
k=0⇒x=π/6
k=1⇒x=π/6+2π=13π/6

-5π/2≤5π/6+2πk≤4π
-15≤5+12k≤24
-20≤12k≤19
-20/12≤k≤19/12
k=-1⇒x=5π/6-2π=-7π/6
k=0⇒x=5π/6
k=1⇒x=5π/6+2π=17π/6

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