Необходимо решить неравенства: √2sinx-1≥0 2cos(2xπ/6)> √3

Решение
1) √2sinx-1≥0
sinx ≥ 1/√2
arcsin(1/√2) + 2πn ≤ x ≤ π - arcsin(1/√2) + 2πn, n∈Z
π/4 + 2πn ≤ x ≤ π - π/4 + 2πn, n∈Z
π/4 + 2πn ≤ x ≤ 3π/4 + 2πn, n∈Z

2)  2cos(2xπ/6)>√3
cos(2xπ/6) > √3/2
- arccos(√3/2) + 2πk < 2xπ/6  < arccos(√3/2) + 2πk, k∈Z
- π/6 + 2πk < 2xπ/6  <  π/6 + 2πk, k∈Z
- 1/2 + 6k < x < 1/2 + 6k, k∈Z
- 1/2 < x < 1/2