tan 5x = 1 untuk 0 ≤ x ≤ 2π
Tentukan himpunan penyelesaian dari persamaan trigonometri berikut!
tan 5x = 1 untuk 0 ≤ x ≤ 2π
Pembahasan:
Kita bisa lakukan perhitungan seperti berikut:
tan 5x = 1 = tan ¼π, sehingga diperoleh:
5x = ¼ π + k . π
x = 1/20 π + k . 1/5 π
x = 1/20 π + k . 4/20 π
- k = 0 → x = 1/20 π + 0 . 4/20 π = 1/20 π
- k = 1 → x = 1/20 π + 1 . 4/20 π = 5/20 π
- k = 2 → x = 1/20 π + 2 . 4/20 π = 9/20 π
- k = 3 → x = 1/20 π + 3 . 4/20 π = 13/20 π
- k = 4 → x = 1/20 π + 4 . 4/20 π = 17/20 π
- k = 5 → x = 1/20 π + 5 . 4/20 π = 21/20 π
- k = 6 → x = 1/20 π + 6 . 4/20 π = 25/20 π
- k = 7 → x = 1/20 π + 7 . 4/20 π = 29/20 π
- k = 8 → x = 1/20 π + 8 . 4/20 π = 33/20 π
- k = 9 → x = 1/20 π + 9 . 4/20 π = 37/20 π
- k = 10 → x = 1/20 π + 10 . 4/20 π = 41/20 π (tidak memenuhi)
Jadi himpunan penyelesaiannya adalah Hp = {1/20, 5/20, 9/20, 13/20, 17/20, 21/20, 25/20, 29/20, 33/20, 37/20}
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