6)>√3...

RockMaster @RockMaster

June 2022 1 7 Report

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

Answers & Comments

nKrynka 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

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