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veronik226798
@veronik226798
September 2021
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Помогите пожалуйста решить!!!!!! ( фото во вложении)
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irinan2014
Verified answer
Решение смотри во вложении.
0 votes
Thanks 1
oganesbagoyan
Verified answer
1)
cos x/8 = (-√2)/2 ;
x/8 = ±(π -π/4) +2π*k , k∈Z.
x =± 6π +16π*k , k∈Z.
2)
sin(3x/5 +π/3) =√3/2 * * * 3x/5 +π/3 =(-1)^k +π*k * * *
a)
3x/5 +π/3 =π/3 + 2π*k , k∈Z ⇒ x =
10π/3*k ,
k∈Z.
b)
3x/5 +π/3 = (π -π/3) +2π*k , k∈Z⇒x = 5π/9 +
10π/3*k
,
k∈Z
3)
-tq(2x -π/6) = 1 ;
tq(2x -π/6) = -1 ;
2x -π/6 = -π/4 +π*k , k∈Z.
x =
-π/24 + π/2*k
,k∈Z.
1 votes
Thanks 1
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Answers & Comments
Verified answer
Решение смотри во вложении.Verified answer
1) cos x/8 = (-√2)/2 ;x/8 = ±(π -π/4) +2π*k , k∈Z.
x =± 6π +16π*k , k∈Z.
2) sin(3x/5 +π/3) =√3/2 * * * 3x/5 +π/3 =(-1)^k +π*k * * *
a) 3x/5 +π/3 =π/3 + 2π*k , k∈Z ⇒ x = 10π/3*k , k∈Z.
b) 3x/5 +π/3 = (π -π/3) +2π*k , k∈Z⇒x = 5π/9 + 10π/3*k , k∈Z
3) -tq(2x -π/6) = 1 ;
tq(2x -π/6) = -1 ;
2x -π/6 = -π/4 +π*k , k∈Z.
x = -π/24 + π/2*k ,k∈Z.