Question

Решите неравенство
а)5tg2x<=3
b)√2sin(п/4+х/2)>=1
с)4sin2xcos2x>=√2​

Answer 1

Ответ:

a)

b)√2sin(π/4+x/2)⩾1

sin(π/4+x/2)⩾1/√2

sin(π/4+x/2)⩾√2/2

π/4+2πk<π/4+x/2<3π/4+2πk

2πk<x/2<π/2+2πk

4πk<x<π+4πk

[4πk;π+4πk]

c)4sin(2x)cos(2x) >= sqrt(2)

2*2sin(2x)cos(2x) >= sqrt(2)

2Sin(4x) >= sqrt(2)

Sin(4x) >= sqrt(2)/2

(3pi/4 + 2n*π) >= 4x >= (π/4 + 2n*π)

π*(3/16 + n/2) <= x >= π(1/16 + n/2)