2asin^2x-2根号3sinxcosx+a+b(a>0)定义域[0,二分之派]值域[-2,-1]求a,b

f(x)=2a+b-acos2x-a√3sin2x
=2a+b-2a(cos2xcosπ/6+sin2xsinπ/6)
=2a+b+2asin(2x+π/6)
a>=0,

2asin²x-2√3asinxcosx+a+b
=2asin²x-√3a(2sinxcosx)+a+b
=a(1-cos2x)-√3 a sin2x+a+b
=-acos2x-√3asin2x+2a+b
=-2a[(1/2)cos2x+√(3/2)sin2x]+2a+b
=-2a[sin2x*cos(π/6)+cos2x*sin(π/6)]+2a+b
=-2asin(2x+π/6)+2a+b
∵x∈[π/2,π],
∴ 2x+π/6∈[7π/6,13π/6]
∴ sin(2x+π/6)∈[-1,1/2]
①a>0时
值域是[a+b,4a+b]
有a+b=2 4a+b=5
得a=b=1
②a

题目是这样的吧:
f（x）=2asin²x-2根号3asinxcosx+a+b,x∈[0,π/2],值域为[-5,1]
2sin^2x=1-cos2x
2sinxcosx=sin2x
f(x)=a-acos2x-a√3sin2x+a+b
=2a+b-2a(cos2xcosπ/3+sin2xsinπ/3)
=2a+b-2acos(2x-π/3)
-5

2asin^x-2根号3asinxcosx+a+b=2asin²x-√3a(2sinxcosx)+a+b=-(a-2asin²x)+a-√3a(2sinxcosx)+a+b=-acos2x-√3asin2x+2a+b=-2a[1/2cos2x+√3/2sin2x]+2a+b=-2asin(2x+π/6)+2a+b又x∈[π/2,π],则2x+π/6∈...

（1）f(x)=a(1-cos2x)-根3*a*sin2x+b
=a(1-cos2x-根3sin2x)+b
=a-2a(1/2cos2x+根3/2sin2x)+b
=a-2asin(π/6+2x)+b
其定义域为[0,π/2] 那么2x值域是[0,π],π/6+2x值域是[π/6,7π/6],那么sin(π/6+2x)的值域为-1/2,1
那么f(x)的值域根据a的符号进行讨论,
当a＞0时,f(x)的值域为b-a,2a+b 此时,a=3,b=-2
当a反折=>尺度变换的顺序进行,例如y=x变成y=-1/2x+1,y=x左移一个单位（平移）然后反折（按Y轴为对称轴）,最后尺度变换（X轴坐标变为原来的2倍,如果X的系数是2,则变为原来的1/2）,想变回去就反着来,如y=-1/2x+1变成y=x 先尺变,再反折,最后平移.