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y000
@y000
December 2021
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Найдите все значения x, при которых 1+2*cos^2 x и (-sinx) равны
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nKrynka
1 + 2*cos^2x = - sinx
1 + 2*cos^2x + sinx = 0
1 + 2*(1 - sin^2x) + sinx = 0
1 + 2 - 2*sin^2x + sinx = 0
2*sin^2x - sinx - 3 = 0
D = 1 + 4*2*3 = 25
1) sinx = (1 - 5)/4
sinx = -1
x = - π/2 + 2πn, n∈Z
2) sinx = (1 + 5)/4
sinx = 3/2 не удовлетворяет условию: I sinx I ≤ 1
2 votes
Thanks 4
y000
Понятно, спасибо за подробный ответ.
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Answers & Comments
1 + 2*cos^2x + sinx = 0
1 + 2*(1 - sin^2x) + sinx = 0
1 + 2 - 2*sin^2x + sinx = 0
2*sin^2x - sinx - 3 = 0
D = 1 + 4*2*3 = 25
1) sinx = (1 - 5)/4
sinx = -1
x = - π/2 + 2πn, n∈Z
2) sinx = (1 + 5)/4
sinx = 3/2 не удовлетворяет условию: I sinx I ≤ 1