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teranova4567
@teranova4567
August 2022
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Помогите решить уравнение срочно!
sin2x = sinx - 2cosx + 1
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nKrynka
Решение
2sinx*cosx = sinx - 2cosx + 1
sinx(2cosx - 1) + (2cosx - 1) = 0
(2cosx - 1)*(sinx + 1) = 0
1) 2cosx - 1 - 0
2cosx = 1
cosx = 1/2
x = (+ --) arccos(1/2) + 2πn, n ∈Z
x =( + -) (π/3) + 2πn, n∈Z
2) sinx + 1 = 0
sinx = -1
x = - π/2 + 2πk, k∈Z
Ответ: x =( + -) (π/3) + 2πn, n∈Z; x = - π/2 + 2πk, k∈Z
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Answers & Comments
2sinx*cosx = sinx - 2cosx + 1
sinx(2cosx - 1) + (2cosx - 1) = 0
(2cosx - 1)*(sinx + 1) = 0
1) 2cosx - 1 - 0
2cosx = 1
cosx = 1/2
x = (+ --) arccos(1/2) + 2πn, n ∈Z
x =( + -) (π/3) + 2πn, n∈Z
2) sinx + 1 = 0
sinx = -1
x = - π/2 + 2πk, k∈Z
Ответ: x =( + -) (π/3) + 2πn, n∈Z; x = - π/2 + 2πk, k∈Z