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NITROLEX
@NITROLEX
July 2022
1
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sin(5/6*pi(6x+1))=cos(1/3*pi(3x+2)). Найти сумму корней уравнения, принадлежащих интервалу (0;1/2)
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oganesbagoyan
Verified answer
Sin( (5/6)*(π(6x+1)) =cos((1/3)*(π(3x+2)) ; x∈(0; 1/2).
---
sin( π*( (5/6)*6x +(5/6)*1) ) =cos( π*((1/3)*3x+(1/3)*2) ) ;
sin( π(5x +5/6)) =cos( π(x+ 2/3) ) ;
sin( π(5x +5/6)) =sin( π/2- π(x+ 2/3) ) ;
sin( π(5x +5/6)) = sin( π(1/2- x- 2/3)
) ;
sin( π(5x +5/6)) = sin(- π(x+1/6) ) ;
sin( π(5x +5/6)) + sin( π(x +1/6)
) =0 ;
2sin( π(3x +1/2))*cos( π
(2x+1/3)) =0 ;
[
sin π(3x +1/2)) =0 ; cos( π
(2x+1/3) )=0
.
а)
π(3x +1/2) =πn ,n∈Z.
3x +1/2 = n ⇒x = -1/6 +n/3 ,если n =1⇒
x =1/6
∈ (0; 1/2) .
* * * 0< -1/6 +n/3 < 1/2
⇔ 1/6<n/3< 1/6+1/2 ⇔1/2<n<2 ⇒n=1* * *
б)
π(2x+1/3) = π/2 +πn ,n∈Z.
2x+1/3 = 1/2 +n ⇒ x =1/12+ n/2,если n =0⇒
x =1/12
∈ (0; 1/2)
.
* * * 0< 1/12 +n/2 < 1/2⇔ - 1/12 <n/2< -1/12+1/2 ⇔-1/6<n<5/6 ⇒n=0* * *
сумма корней будет: (1/6 +1/12) =1/4.
ответ :
1/4 .
2 votes
Thanks 1
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Answers & Comments
Verified answer
Sin( (5/6)*(π(6x+1)) =cos((1/3)*(π(3x+2)) ; x∈(0; 1/2).---
sin( π*( (5/6)*6x +(5/6)*1) ) =cos( π*((1/3)*3x+(1/3)*2) ) ;
sin( π(5x +5/6)) =cos( π(x+ 2/3) ) ;
sin( π(5x +5/6)) =sin( π/2- π(x+ 2/3) ) ;
sin( π(5x +5/6)) = sin( π(1/2- x- 2/3) ) ;
sin( π(5x +5/6)) = sin(- π(x+1/6) ) ;
sin( π(5x +5/6)) + sin( π(x +1/6) ) =0 ;
2sin( π(3x +1/2))*cos( π(2x+1/3)) =0 ;
[ sin π(3x +1/2)) =0 ; cos( π(2x+1/3) )=0 .
а)
π(3x +1/2) =πn ,n∈Z.
3x +1/2 = n ⇒x = -1/6 +n/3 ,если n =1⇒ x =1/6 ∈ (0; 1/2) .
* * * 0< -1/6 +n/3 < 1/2⇔ 1/6<n/3< 1/6+1/2 ⇔1/2<n<2 ⇒n=1* * *
б)
π(2x+1/3) = π/2 +πn ,n∈Z.
2x+1/3 = 1/2 +n ⇒ x =1/12+ n/2,если n =0⇒ x =1/12 ∈ (0; 1/2).
* * * 0< 1/12 +n/2 < 1/2⇔ - 1/12 <n/2< -1/12+1/2 ⇔-1/6<n<5/6 ⇒n=0* * *
сумма корней будет: (1/6 +1/12) =1/4.
ответ : 1/4 .