Tentukan Himpunan Penyelesaian dari sin(3x + (π/6)) = ½√2 untuk 0 ≤ x ≤ 2π

Jawaban yang benar adalah {π/36, 7π/36, 25π/36, 31π/36, 49π/36, 55π/36} Ingat kembali: Persamaan trigonometri untuk sinus: sin x = sin α x1 = α + k·2π x2 = (π - α) + k·2π dengan k ∈bilangan bulat Pembahasan: sin {3x + (π/6)} = ½√2 sin {3x + (π/6)} = sin π/4 3x1 + (π/6) = (π/4) + k·2π 3x1 = {(π/4) - (π/6)} + k·2π 3x1 = {(3π - 2π)/12} + k·2π 3x1 = (π/12) + k·2π x1 = {(π/12)/3} + k·(2π/3) x1 = (π/36) + k·(2π/3) k = 0 x1 = (π/36) + 0·(2π/3) x1 = (π/36) + 0 x1 = π/36 (memenuhi interval) k = 1 x1 = (π/36) + 1·(2π/3) x1 = (π/36) + (2π/3) x1 = (π + 24π)/36 x1 = 25π/36 (memenuhi interval) k = 2 x1 = (π/36) + 2·(2π/3) x1 = (π/36) + (4π/3) x1 = (π + 48π)/36 x1 = 49π/36 (memenuhi interval) k = 3 x1 = (π/36) + 3·(2π/3) x1 = (π/36) + 2π x1 = (π + 72π)/36 x1 = 73π/36 (tidak memenuhi interval) 3x2 + (π/6) = {π - (π/4)} + k·2π 3x2 + (π/6) = (3π/4) + k·2π 3x2 = {(3π/4) - (π/6)} + k·2π 3x2 = {(9π - 2π)/12} + k·2π 3x2 = (7π/12) + k·2π x2 = {(7π/12)/3} + k·(2π/3) x2 = (7π/36) + k·(2π/3) k = 0 x2 = (7π/36) + 0·(2π/3) x2 = (7π/36) + 0 x2 = 7π/36 (memenuhi interval) k = 1 x2 = (7π/36) + 1·(2π/3) x2 = (7π/36) + (2π/3) x2 = (7π + 24π)/36 x2 = 31π/36 (memenuhi interval) k = 2 x2 = (7π/36) + 2·(2π/3) x2 = (7π/36) + (4π/3) x2 = (7π + 48π)/36 x2 = 55π/36 (memenuhi interval) k = 3 x2 = (7π/36) + 3·(2π/3) x2 = (7π/36) + 2π x2 = (7π + 72π)/36 x2 = 79π/36 (tidak memenuhi interval) Jadi, himpunan penyelesaian dari persamaan trigonometri di atas adalah {π/36, 7π/36, 25π/36, 31π/36, 49π/36, 55π/36}