Ответы на вопрос:
Решение log₂ sin(x/2) < - 1 одз: sinx/2 > 0 2πn < x/2 < π + 2πn, n ∈ z 4πn < x < 2π + 4πn, n ∈ z sin(x/2) < 2⁻¹ sin(x/2) < 1/2 - π - arcsin(1/2) + 2πn < x/2 < arcsin(1/2) + 2πn, n ∈ z - π - π/6 + 2πn < x/2 < π/6 + 2πn, n ∈ z - 7π/6 + 2πn < x/2 < π/6 + 2πn, n ∈ z - 7π/3 + 4πn < x < π/3 + 4πn, n ∈ z 2) log₁/₂ cos2x > 1 одз: cos2x > 0 - arccos0 + 2πn < 2x < arccos0 + 2πn, n ∈ z - π/2 + 2πn < 2x < π/2 + 2πn, n ∈ z - π + 4πn < x < π + 4πn, n ∈ z так как 0 < 1/2 < 1, то cos2x < 1/2 arccos(1/2) + 2πn < 2x < 2π - arccos(1/2) + 2πn, n ∈ z π/3 + 2πn < 2x < 2π - π/3 + 2πn, n ∈ z π/6 + πn < x < 5π/6 + πn, n ∈ z
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