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
Ответы
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.