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Aseme
@Aseme
July 2022
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Решите неравенство: sinx + cos2x > 1.
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nKrynka
Решение
sinx + cos2x > 0
sinx + 1 - 2sin^2(x) > 0
2sin^2(x) - sinx - 1 < 0
sinx = z
2*z^2 - z - 1 = 0
D = 1 + 4*1*1 = 9
z1 = (1 - 3)/4
z1 = -1/2
z2 = (1 + 3)/4
z2 = 1
1) sinx = -1/2
x = (-1)^(n)arcsin(-1/2) + πn, n∈Z
x = (-1)^(n+1)arcsin(1/2) + πn, n∈Z
x1 = (-1)^(n+1)(π/6) + πn, n∈Z
2) sinx = 1
x2 = π/2 = 2πk, k∈Z
Ответ: x1 = (-1)^(n+1)(π/6) + πn, n∈Z; x2 = π/2 = 2πk, k∈Z.
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Answers & Comments
sinx + cos2x > 0
sinx + 1 - 2sin^2(x) > 0
2sin^2(x) - sinx - 1 < 0
sinx = z
2*z^2 - z - 1 = 0
D = 1 + 4*1*1 = 9
z1 = (1 - 3)/4
z1 = -1/2
z2 = (1 + 3)/4
z2 = 1
1) sinx = -1/2
x = (-1)^(n)arcsin(-1/2) + πn, n∈Z
x = (-1)^(n+1)arcsin(1/2) + πn, n∈Z
x1 = (-1)^(n+1)(π/6) + πn, n∈Z
2) sinx = 1
x2 = π/2 = 2πk, k∈Z
Ответ: x1 = (-1)^(n+1)(π/6) + πn, n∈Z; x2 = π/2 = 2πk, k∈Z.