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