4sin2xcoc2x>или= корень 2

Question

4sin2xcoc2x>или= корень 2

Answer 1

4sin2x*cos2x ≥ √2
2(2sin2x*cos2x) ≥ √2
2*sin4x ≥ √2
sin4x ≥ √2/2
π/4+2πn ≤ 4x ≤ 3π/4+2πn, n∈Z
π/16+πn/2 ≤ x ≤ 3π/16 +πn/2, n∈Z
x∈[π/16+πn/2; 3π/16 +πn/2], n∈Z