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hurume
@hurume
June 2022
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решите пожалуйста первый варинт
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Кулшат
1) √2cos4x = 1; cos4x = √2/2; 4x = +-π/4+2πn, n∈Z; x =+- π/14 +πn/2,n∈Z;
2) (π/3 - 3x) = +- π/3 + 2πn,n∈Z; 3x = +-π/3 - π/3 + 2πn ,n∈Z;
x =+-π/9 - π/9 + 2πn/3, n∈Z.
3) x = π/2 +πn,n∈Z.
4) sin(2x+π/6 + x) =1/2; 3x+π/6 = (- 1)^{n} π/6 +πn,n∈Z;
3x = (- 1)^{n} π/6 -π/6 +πn,n∈Z; x= (- 1)^{n} π/18+πn/3. n∈Z.
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hurume
3 это задание с косинусом или котангенсом?
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Answers & Comments
2) (π/3 - 3x) = +- π/3 + 2πn,n∈Z; 3x = +-π/3 - π/3 + 2πn ,n∈Z;
x =+-π/9 - π/9 + 2πn/3, n∈Z.
3) x = π/2 +πn,n∈Z.
4) sin(2x+π/6 + x) =1/2; 3x+π/6 = (- 1)^{n} π/6 +πn,n∈Z;
3x = (- 1)^{n} π/6 -π/6 +πn,n∈Z; x= (- 1)^{n} π/18+πn/3. n∈Z.