y=cos^2x+sinx+1(|x|

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y=cos^2x+sinx+1(|x|
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y=cos^2x+sinx+1(|x|
y=cos^2x+sinx+1(|x|

y=cos^2x+sinx+1(|x|
|x|<=π/3
则-π/3<=x<=π/3
-√3/2<=sinx<=√3/2
y=cos^2x+sinx+1
=1-sin^2x+sinx+1
=-(sin^2x-sinx+1/4)+1/4+2
=-(sinx-1/2)^2+9/4
sinx=-√3/2时,Ymin=-1-√3/2+9/4=5/4-√3/2
sinx=1/2时,Ymax=9/4
所以,所求值域:
5/4-√3/2<=y<=9/4

y=cos²x+sinx+1=1-sin²x+sinx+1=-sin²x+sinx+2=-(sinx-1/2)²+9/4
-∏/3≤x≤∏/3
-√3/2≤sinx≤√3/2
sinx=1/2时有最大值,y=9/4
sinx=-√3/2有最小值,y=(5-2√3)/4
y的值域[(5-2√3)/4,9/4]
sinx 在[-√3/2,1/2]单调递增

设sinx=X,-√3/2<=X<=√3/2
那么y=2+X-2X^2=9/4-2(1/2+X)^2
最大为X=-1/2,y=9/4
最小为X=√3/2,y=1/2+√3/2