Question

ПОМОГИТЕ ПОЖАЛУЙСТА срочно 22.54(б,в)

Answer 1

cos(π/4 - x) < √2/2

π/4 - 2πk < π/4 - x < 3π/4 - 2πk
- 2πk < -x < π/2 - 2πk
-π/2 + 2πk < x < 2πk, k∈Z

в) 
cos(3x -π/6) > -1/4; 0 ≤ arccos(-1/4) ≤ π
arccos(-1/4) + 2πk < 3x - π/6 < 2π - arccos(-1/4) +2πk
π/6 + arccos(-1/4) + 2πk < 3x < 2π + π/6 - arccos(-1/4) +2πk
π/18 + arccos(-1/4)/3 + 2πk/3 < x < 2π/3 + π/18 - arccos(-1/4)/3 +2πk/3, k∈Z