設 y = (2+cosx)/(2+sinx)
y(2+sinx) = 2+cosx
2y + y sinx = 2 + cosx
y sinx - cosx = 2 - 2y
由於 - √[y²+1] ≤ y sinx - cosx ≤ √[y²+1]
因此 - √[y²+1] ≤ 2 - 2y ≤ √[y²+1]
(2-2y)² ≤ y²+1
4y² - 8y + 4 ≤ y² + 1
3y² - 8y + 3 ≤ 0
[8 - √(64-36)] / 6 ≤ y ≤ [8 + √(64-36)] / 6
[8 - √28] / 6 ≤ y ≤ [8 + √28] / 6
[4 - √7]/3 ≤ y ≤ [4 + √7]/3
