Jawab:
[1.]  Hp = {⅓ π,  ⁵/₆ π,  ⁴/₃ π}
[2.] Hp = {90°, 210°}

Penjelasan:
[1.]  tan(2x + ¹/₆π) = -⅓√3
dengan daerah asal {0 ≤ x ≤ ³/₂}
2x + ¹/₆π = tan⁻¹ -⅓√3

nilai tan⁻¹ negatif pd kuad ii dan iv
pd kuad ii, sudut x = π - tan⁻¹(x)
pd kuad iv, sudut x = 2π - tan⁻¹(x)
Periodisitas tan x = π, maka + k · π
Nilai dari tan⁻¹ -⅓√3 = ¹/₆π rad.

Kuad. ii.
2x + ¹/₆ π = tan⁻¹ -⅓√3 + k · π
2x + ¹/₆ π = π - (tan⁻¹ ⅓√3) + k · π
2x + ¹/₆ π = π - (¹/₆ π) + k · π
2x + ¹/₆ π = (⁶/₆ π - ¹/₆ π) + k · π
2x + ¹/₆ π = ⁵/₆ π + k · π
2x = (⁵/₆ π - ¹/₆ π) + k · π
2x = ⁴/₆ π + k · π
2x = ⅔ π + k · π
(2x ÷ 2) = ⅔₍₂₎ π + k · ½π
x = ⅓ π + k · ½π
karena daerah asal 0 ≤ x ≤ ³/₂, k ≠ 3
k = 0, x = ⅓ π + 0 · ½π = ⅓ π
k = 1, x = ⅓ π + 1 · ½π = ⅓ π + ½π = ²/₆ π + ³/₆ π = ⁵/₆ π
k = 2, x = ⅓ π + 2 · ½π = ⅓ π + π = ⅓ π + ³/₃π = ⁴/₃ π

Kuad. iv.
2x + ¹/₆ π = tan⁻¹ -⅓√3 + k · π
2x + ¹/₆ π = 2π - (tan⁻¹ ⅓√3) + k · π
2x + ¹/₆ π = 2π - (¹/₆ π) + k · π
2x + ¹/₆ π = (¹²/₆ π - ¹/₆ π) + k · π
2x + ¹/₆ π = ¹¹/₆ π + k · π
2x = (¹¹/₆ π - ¹/₆ π) + k · π
2x = ¹⁰/₆ π + k · π
2x = ⁵/₃ π + k · π
(2x ÷ 2) = ⁵/₃₍₂₎ π + k · ½π
x = ⁵/₆ π + k · ½π
karena daerah asal 0 ≤ x ≤ ³/₂, k ≠ 3
k = 0, x = ⁵/₆ π + 0 · ½π = ⁵/₆ π
k = 1, x = ⁵/₆ π + 1 · ½π = ⁵/₆ π + ½π = ⁵/₆ π + ³/₆ π = ⁸/₆ π = ⁴/₃ π
k = 2, x = ⁵/₆ π + 2 · ½π = ⁵/₆ π + π = ⁵/₆ π + ⁶/₆ π = ¹¹/₆ π (tdk memenuhi)

Hp = {⅓ π,  ⁵/₆ π,  ⁴/₃ π}

[2.] sin x - √3 cos x - 1 = 0
Daerah asal {0° ≤ x ≤ 360°}
sin x - 1 = √3 cos x
(sin x - 1)² = (√3 cos x)²
sin²x - 2sin x + 1 = 3 cos²x
∵ cos² x = 1 - sin² x ∴
sin²x - 2sin x + 1 = 3 cos²x
sin²x - 2sin x + 1 = 3 (1 - sin² x)
sin²x - 2sin x + 1 = 3 - 3sin² x
sin²x + 3sin² x - 2sin x + 1 - 3 = 0
4sin² x - 2sin x - 2 = 0
⁴/₂ sin² x - ²/₂ sin x - ²/₂ = 0
2sin² x - sin x - 1 = 0
sin² x - sin x - 2 = 0
½(2sin x + 1)(2sin x - 2) = 0
(2sin x + 1)(sin x - 1) = 0
sin x = -½    sin x = 1
x = sin⁻¹ -½       x = sin⁻¹ 1

Cari Hp jika x = sin⁻¹ -½
nilai sin⁻¹ negatif pd kuad iii dan iv
maka sin⁻¹ -½ = 180° + sin⁻¹ ½ = 360° - sin⁻¹ ½
∵ sin⁻¹ ½ = 30° ∴
sin⁻¹ -½ = 180° + 30° = 360° - 30°
sin⁻¹ -½ = 210° = 330°
sin x periodisitasnya 360°, maka
tidak ada solusi lain selain 210°dan330°

Cari Hp jika x = sin⁻¹ 1
nilai sin⁻¹ positif pada kuad i dan ii
maka sin⁻¹ 1 = 180° - sin⁻¹ 1
∵ sin⁻¹ 1 = 90° ∴
sin⁻¹ 1 = 90° = 180° - 90°
sin⁻¹ 1 = 90°
sin x periodisitasnya 360°, maka
tidak ada solusi lain selain 90°

Nilai x sementara {90°, 210°, 330°}

Cek solusi, untuk x = 90°
sin x - √3 cos x - 1 = 0
sin 90° - √3 cos 90° - 1 = 0
1 - √3(0) - 1 = 0
1 - 1 = 0 [memenuhi]   x = 90°

Cek solusi, untuk x = 210°
sin x - √3 cos x - 1 = 0
sin 210° - √3 cos 210° - 1 = 0
sin (180+30)° - √3 cos(180+30)° - 1 = 0
-(sin30°) - √3 (-cos30°) - 1 = 0
-½ - √3 (-½√3) - 1 = 0
-½ + ³/₂ - 1 = 0
²/₂ - 1 = 0
1 - 1 = 0 [memenuhi]   x = 210°

Cek solusi, untuk x = 330°
sin x - √3 cos x - 1 = 0
sin 330° - √3 cos 330° - 1 = 0
sin (360-30)° - √3 cos(360-30)° - 1 = 0
-(sin30°) - √3 (cos30°) - 1 = 0
-½ - √3 (½√3) - 1 = 0
-½ - ³/₂ - 1 = 0
-⁴/₂ - 1 = 0
-2 - 1 = 0
-3 ≠ 0 [tidak memenuhi]

Hp = {90°, 210°}

(xcvi)