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Артур998
@Артур998
October 2021
1
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Решите номер 7.Есть вложение.
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oganesbagoyan
Verified answer
7.
sin(x/2) *(cosx+1) =0⇔ [sin(x/2) =0 ; cosx+1=0⇔ [sin(x/2) =0 ; cosx =-1.⇔
[x/2 =πn ; x = π +2πn ,n∈Z.⇔ [x =2πn ; x = π +2πn ,n∈Z.⇔
[
x =π*(2n) ; x = π*(2n+1) ,n∈Z
⇔
x =πk , k∈Z
.
1 votes
Thanks 0
oganesbagoyan
НЕ НАДО ! ФУНКЦИИ SIN И COS ОПРЕДЕЛЕНЫ x∈ (-∞;∞).
oganesbagoyan
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Answers & Comments
Verified answer
7.sin(x/2) *(cosx+1) =0⇔ [sin(x/2) =0 ; cosx+1=0⇔ [sin(x/2) =0 ; cosx =-1.⇔
[x/2 =πn ; x = π +2πn ,n∈Z.⇔ [x =2πn ; x = π +2πn ,n∈Z.⇔
[ x =π*(2n) ; x = π*(2n+1) ,n∈Z
⇔x =πk , k∈Z.