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Yukka16
@Yukka16
July 2022
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Найдите корни уравнения 2sin ( x-1)= -корень из 2, принадлежащие промежутку [0; 2пи]
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mikael2
Verified answer
2sin(x-1)= -√2 sin(x-1)= -√2/2
x-1=(-1)ⁿ(-π/4)+πn
x=(-1)ⁿ⁺¹π/4+πn+1 n∈Z
x∈[0;2π] n=0 y=
-π/4+1
≈-0.8+1=0.2
n= -1 y=π/4 -π+1=-3/4π+1≈-2.3+1<0
n=1 y=π/4+π+1=
5π/4+1
<2π
n=2 y=-π/4+2π+1 1-π/4>0 так как -π/4≈-0,8 и у>2π
n= -2 y= -π/4-2π+1<0
ответ
1-π/4; 1+5π/4
4 votes
Thanks 3
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Answers & Comments
Verified answer
2sin(x-1)= -√2 sin(x-1)= -√2/2x-1=(-1)ⁿ(-π/4)+πn
x=(-1)ⁿ⁺¹π/4+πn+1 n∈Z
x∈[0;2π] n=0 y=-π/4+1≈-0.8+1=0.2
n= -1 y=π/4 -π+1=-3/4π+1≈-2.3+1<0
n=1 y=π/4+π+1=5π/4+1<2π
n=2 y=-π/4+2π+1 1-π/4>0 так как -π/4≈-0,8 и у>2π
n= -2 y= -π/4-2π+1<0
ответ 1-π/4; 1+5π/4