A. maths 高手請進~!

Let θ/2 be x.
2cosθ + sinθ - 1 = 0
2cos2x + sin2x - 1 = 0
2(cos²x - sin²x) +2 sinx cosx - (cos²x + sin²x) = 0
cos²x + 2sinx cosx - 3 sin²x = 0
(cosx + 3sinx) (cosx - sinx) = 0
cosx = -3sinx or cosx =sinx
tan x = 1 or -1/3

Since α and β are roots, WLOG, tan β/2 =1 and tan α/2 = -1/3

Let y = α/2, then tan y = -1/3 and let z = β/2, then tan z = 1

tan(α+β)/2 = tan(y+z) = (tan y + tan z) / [ 1- tan y tan z]
=[(-1/3) + 1] / [1-(-1/3)(1) ] = ½

tan(α+β) = tan 2[ (α+β)/2 ] = 2tan [ (α+β)/2 ] / [ 1 - tan² (α+β)/2 ]
=2(½) / [1-(½)²] = 4/3

你可唔可以打中文我唔係好識英文 ,但係我可以教你用輔助角
first, you can畫一個3個形
之後就cosα cosx+sinα sinx=1/開方3
cos(α -x)=1/開方3,其中α 是tanα =1/開方3