nilai x yang memenuhi persamaan trigonometri akar 6 sin x – akar 2 cos x = -2 untuk 0 < x < 2 phi

A cos x + b sin x = k cos(x – @) dengan k = √(a^2 + b^2) dan tan @ = b/a
√6 sin x – √2 cos x = -2
=> -√2 cos x + √6 sin x = -2
a = -√2 dan b = √6 => kuadran 2

k = √((-√2)^2 + (√6)^2) = √(2 + 6) = √8 = 2√2
tan @ = b/a = √6/(-√2) = -√3 => @ = 120°

-√2 cos x + √6 sin x = -2
=> 2√2 cos (x – 120°) = -2
=> cos (x – 120°) = -2/(2√2)
=> cos (x – 120°) = (-1/2)√2
=> cos (x – 120°) = cos 135°
=> x – 120 = 135 + k.360 atau x – 120 = -135 + k.360
=> ……… x = 255 + k.360 ……………… x = -15 + k.360
k = 0 => x = 255………..………………….. x = -15 (tdk memenuhi/TM)
k = 1 => x = 615 (TM) ……………………. x = 345
x = 255° = 255/180 π = (17/12)π
x = 345° = 345/180 π = (23/12)π