Question

решите неравенства:4sin2xcox2x > корня из 3
2 cos(4x + pi/3) < корня из 3

Answer 1

4sin2xcos2x > √3
2sin4x > √3
sin4x > √3/2
π/3 + 2πn < 4x < 2π/3 + 2πn
π/12 + πn/2 < x < π/6 + πn/2

ОТВЕТ: (π/12 + πn/2; π/6 + πn/2) n∈Z

2cos(4x + π/3) < √3
cos(4x + π/3) < √3/2
π/6 + 2πn < 4x + π/3 < 11π/6 + 2πn
π/6 - π/3 + 2πn < 4x < 11π/6 - π/3 + 2πn
-π/6 + 2πn < 4x < 3π/2 + 2πn
-π/24 + πn/2 < x < 3π/8 + πn/2

ОТВЕТ: (-π/24 + πn/2; 3π/8 + πn/2) n∈Z