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abieva
@abieva
August 2022
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помогите очень срочно нужно
[tex]sin( \pi cos3x)=1[/tex]
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irka1804
Sin(πcos3x) = 1
πcos3x = π/2 + 2πn, n ∈ Z
cos3x = 1/2 + 2n, n ∈ Z
-1 ≤ cos3x ≤ 1
-1 ≤ 1/2 + 2n ≤ 1
-3/2 ≤ 2n ≤ 1/2
n = 0
cos3x = 1/2
3x = +- π/3 + 2πk, k ∈ Z
x = +- π/9 + 2πk / 3, k ∈ Z
1 votes
Thanks 2
abieva
ой, я уже решила ;) спасибо
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Answers & Comments
πcos3x = π/2 + 2πn, n ∈ Z
cos3x = 1/2 + 2n, n ∈ Z
-1 ≤ cos3x ≤ 1
-1 ≤ 1/2 + 2n ≤ 1
-3/2 ≤ 2n ≤ 1/2
n = 0
cos3x = 1/2
3x = +- π/3 + 2πk, k ∈ Z
x = +- π/9 + 2πk / 3, k ∈ Z