Помогите, пожалуйста! 1. log2 sin x/2 < -1 2. log1/2 cos 2x > 1

Решениеlog₂ sin(x/2) < - 1ОДЗ: sinx/2 > 02πn < x/2 < π + 2πn, n ∈ Z4πn < x < 2π + 4πn, n ∈ Zsin(x/2) < 2⁻¹sin(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 ∈ Z2)  log₁/₂ cos2x > 1ОДЗ:cos2x > 0- arccos0 + 2πn < 2x < arccos0 + 2πn, n ∈ Z- π/2 + 2πn < 2x < π/2 + 2πn, n ∈ Z- π + 4πn < x < π + 4πn, n ∈ Zтак как 0 < 1/2 < 1, тоcos2x < 1/2arccos(1/2) + 2πn < 2x < 2π - arccos(1/2) + 2πn, n ∈ Zπ/3 + 2πn < 2x < 2π - π/3 + 2πn, n ∈ Zπ/6 + πn < x < 5π/6 + πn, n ∈ Z