tentukan himpunan penyelesian persamaan 3 tan ( 2x + 2/3 π ) = √3 pada interval 0 < x < π tolong bantuin dong ,, makasih

Bab Trigonometri
Matematika SMA Kelas X

0 < x < π → kuadran I dan II

3 tan (2x + 2π/3) = √3
tan (2x + 2π/3) = (√3)/3

2x + (2π/3) = π/6
2x = π/6 - 2π/3 → minus

2x + (2π/3) = 7π/6
2x = 7π/6 - 2π/3
2x = 7π/6 - 4π/6
x = 3π/(2 . 6)
x = π/4

2x + (2π/3) = 13π/6
2x = 13π/6 - 2π/3
2x = 13π/6 - 4π/6
2x = 9π/6
x = 9π/(2 . 6)
x = 3π/4

HP = { π/4, 3π/4 }

3 tan (2x + 2/3 π) = √3
tan (2x + 2/3 π) = 1/3 √3
tan (2x + 2/3 π) = tan π/6
2x + 2/3 π = π/6 + k . π
2x = π/6 - 2/3 π + k . π ====> π/6 - 4/6 π = -3/6 π = -1/2 π
2x = -1/2 π + k . π
x = -1/4 π + k . 1/2 π
k = 0 => x = -1/4 π (tidak memenuhi)
k = 1 => x = 1/4 π
k = 2 => x = 3/4 π
k = 3 => x = 5/4 π (tidak memenuhi)
HP = {1/4 π, 3/4 π} = {π/4, 3π/4}