Question

Tentukan HP dari (dengan cara):
1. Cos 4x = sin x
dengan 0 ≤ x ≤ 2

2. Sin 2x = cos x
dengan 0 ≤ x ≤ 2

Answer 1

Nomor 1

cos 4x = sin x
cos 4x = cos (1/2π - x)
       4x = 1/2π - x      + k.2π ....(1)
       4x = -(1/2π - x) + k.2π ........(2)

Persamaan (1)
4x = 1/2π - x + k.2π
5x = 1/2π + k.2π
  x = 1/10π + k.2/5π
  x = 1/10π + k.4/10π

k = 0 → x = 1/10π
k = 1 → x = 1/10π + 1.4/10π = 1/2π
k = 2 → x = 1/10π + 2.4/10π = 6/10π
k = 3 → x = 1/10π + 3.4/10π = 13/10π
k = 4 → x = 1/10π + 4.4/10π = 17/10π

Persamaan (2)
4x = -(1/2π - x) + k.2π
4x = -1/2π + x + k.2π
3x = -1/2π + k.2π
 x = -1/6π + k.2/3π
 x = -1/6π + k.4/6π

k=1 → x = -1/6π + 1.4/6π = 1/2π
k=2 → x = - 1/6π + 2.4/6π = 7/6π
k=3 → x = -1/6π + 3.4/6π = 11/6π

Hp = {1/10π, 1/2π, 9/10π, 7/6π, 13/10π, 17/10π, 11/6π}

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Nomor 2

sin 2x = cos x
sin 2x = sin(1/2π - x)
     2x = (1/2π - x) + k.2π ....(1)
     2x = [π - (1/2π - x) + k.2π ....(2)

Persamaan (1)
2x = 1/2π - x + k.2π
  x = 1/6π + k.2/3π
  x = 1/6π + k.4/6π

k = 0 → x = 1/6π
k = 1 →x = 1/6π + 1.4/6π = 5/6π
k = 2 → x = 1/6π + 2.4/6π = 9/6π

Persamaan (2)
     2x = [π - (1/2π - x)] + k.2π
         = 1/2π + x + k.2π
     x = 1/2π + k.2π

k = 0 → x = 1/2π

Hp = {1/6π, 1/2π, 5/6π, 9/6π }