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koshelevva
@koshelevva
July 2022
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cos2x=Корень2(cosx-sinx)
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Cos²x - sin²x - √2(cosx - sinx) = 0
(cosx - sinx)(cosx + sinx) - √2(cosx - sinx) = 0
(cosx - sinx) · (cosx + sinx - √2) = 0
cosx - sinx = 0 или cosx + sinx - √2 = 0
1 - tgx = 0 1/√2 · cosx + 1/√2 · sinx = 1
tgx = 1 sin(x + π/4) = 1
x = π/4 + πn x + π/4 = π/2 + 2πk
x = π/4 + 2πk
Вторая группа корней включается в первую, поэтому
Ответ: π/4 + πn
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Answers & Comments
Verified answer
Cos²x - sin²x - √2(cosx - sinx) = 0(cosx - sinx)(cosx + sinx) - √2(cosx - sinx) = 0
(cosx - sinx) · (cosx + sinx - √2) = 0
cosx - sinx = 0 или cosx + sinx - √2 = 0
1 - tgx = 0 1/√2 · cosx + 1/√2 · sinx = 1
tgx = 1 sin(x + π/4) = 1
x = π/4 + πn x + π/4 = π/2 + 2πk
x = π/4 + 2πk
Вторая группа корней включается в первую, поэтому
Ответ: π/4 + πn