已知关于x的方程cos^2x-sin^2x+2sinx+2a+1=0在区间〔0,π/2]内有解,则实数a的取值范围是

〔0,π/2]
cos^2x-sin^2x+2sinx+2a+1=0
1-sin^2x-sin^2x+2sinx+2a+1=0
sin^2x-sinx-a-1=0
(sinx-1/2)^2=a+5/4
sinx=1/2[1±根号(4a+5]
x∈〔0,π/2]
0＜sinx≤1
0＜1/2[1±根号(4a+5]≤1
-1＜±根号(4a+5)≤1
由 -1＜根号(4a+5)≤1 得：-5/4 ≤ a ≤ -1
由 -1＜-根号(4a+5)≤1 得：-5/4 ≤ a ＜ -1
综上：-5/4 ≤ a ＜ -1

化简一下： 1-3sin^2x 2sinx 2a 1=0 3sin^2x-2sinx-2a-2=0 sin^2x0