Помогите решить:

tg (π/4 - 2x) ≥1

2cos (x+π/3) - √3 < 0

sin2x / (1 + sinx) = - 2cosx

1-cosx=sin x/2

√3/2 sin3x - (1) / 2 cos3x=-1

sin (x) + sin (2x) + sin (3x) = cos (x) + cos (2x) + cos (3x)

sin (2x) + sin (2x - x) + sin (2x + x) = cos (2x) + cos (2x - x) + cos (2x + x)

sin (2x) + sin (2x) ·cos (x) - cos (2x) ·sin (x) + sin (2x) ·cos (x) + cos (2x) ·sin (x) =

= cos (2x) + cos (2x) ·cos (x) + sin (2x) ·sin (x) + cos (2x) ·cos (x) - sin (2x) ·sin (x)

sin (2x) + 2·sin (2x) ·cos (x) = cos (2x) + 2·cos (2x) ·cos (x)

sin (2x) ·[1 + 2·cos (x) ] = cos (2x) ·[1 + 2·cos (x) ]

[sin (2x) - cos (2x) ]·[1 + 2·cos (x) ] = 0

1) sin (2x) - cos (2x) = 0

sin (2x) = cos (2x)

tg (2x) = 1

2x = π/4 + π·n = π (4n + 1) / 4

x = π (4n + 1) / 8

2) 1 + 2·cos (x) = 0

cos (x) = - ½

x = ±2π/3 + 2·π·n = 2π (3n ± 1) / 3

Ответ:

{x = π (4n + 1) / 8

{x = 2π (3n ± 1) / 3

n - целое.