Question

1. 2 cos x-1 ≥0 решите #305

Answer 1

Решение:

1) 2 cos x-1 ≥ 0

cosx ≥ 1/2

- arccos (1/2) + 2πn ≤ x ≤ arccos (1/2) + 2πn, n ∈ Z

- π/3 + 2πn ≤ x ≤ π/3 + 2πn, n ∈ Z2) 2sinx + √2 ≥ 0

sinx ≥ - √2/2

arcsin(√2/2) + 2πn ≤ x ≤ π - arcsin(√2/2) + 2πn, n ∈ Z

π/4  + 2πn ≤ x ≤ π - π/4 + 2πn, n ∈ Z

π/4  + 2πn ≤ x ≤  3π/4 + 2πn, n ∈ Z

3) 2cosx - √3 ≤ 0

2cosx ≤ √3

cosx ≤ √3/2

π/6 + 2πn ≤ x ≤  2π - π/6 + 2πn, n ∈ Z

π/6 + 2πn ≤ x ≤  11π/6 + 2πn, n ∈ Z

4)  3tgx + √3 > 0

tgx > - √3/3

arctg(- √3/3) + πn ≤ x ≤ π/2 + πn, n ∈ Z

- π/6 + πn ≤ x ≤ π/2 + πn, n ∈ Z

Answer 2

Спасибо большое