if k is an interger which

令x=θ
2sin^2 x + 10cos^2 (x/2) = 7-2k
2sin^2 x + 5(1+cos x) = 7-2k (半角公式)
2(1- cos^2 x)+5+5 cos x = 7-2k
-2cos^2 x +5cos x +7 = 7-2k
-2cos^2 x +5cos x = -2k
cos^2 x - (5/2)cos x = k
(cos x - 5/4)^2 - 25/16 = k

-1 <= cos x <= 1
-9/4 <= cos x - 5/4 <= -1/4
1/16 <= (cos x -5/4)^2 <= 81/16
-24/16 <= (cos x - 5/4)^2 - 25/16 <= 56/16
-3/2 <= k <= 7/2

k = -1, 0, 1, 2, 3 故選(B)


2013-07-10 09:33:06 補充：
意見欄螞蟻雄兵知識長的答案為(B)(C)(E)，我想答案一樣為何選項不一樣，仔細一看，原來題目是k"屬於"什麼集合‧字太小我看成k"等於"什麼值，既然是"屬於"什麼集合，答案不變，選項要修正為(B)(C)(E)。

Sol
2Sin^2 θ+10Cos^2 (θ/2)=7－2k
2k=7－2Sin^2 θ－10Cos^2 (θ/2)
k=3.5－Sin^2 θ－5Cos^2 (θ/2)
=3.5－1+Cos^2 θ－2.5(1+Cosθ)
=Cos^2 θ－2.5Cosθ
=Cos^2 θ－2.5Cosθ+1.25^2－1.25^2
=(Cosθ－1.25)^2－1.5625

2013-07-10 09:14:11 補充：
－1<=Cosθ<=1
－2.25<=Cosθ－1.25<=－0.25
0.0625<=(Cosθ－1.25)^2<=5.0625
0.0625－1.5625<=(Cosθ－1.25)^2－1.5625<=5.0625－1.5625
－1.5<=k<=3.5
－1<=k<=3
(B)，(C)，(E)

William大師解很好
讚

關鍵在換成同一三角函數
其中要用到半角公式和基本變換公式
最後再用配方法和值範圍求