求函数y=cos(x/2+π/3),x属于[-2π,2π]的单调递减区间
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答：
y=cos(x/2+π/3)
求导：
y'(x)=-sin(x/2+π/3)*(1/2)
=-(1/2)*sin(x/2+π/3)
因为：-2π<=x<=2π
所以：-π<=x/2<=π,-2π/3<=x/2+π/3<=4π/3
所以：-1<=sin(x/2+π/3)<=1
解y'(x)=-(1/2)sin(x/2+π/3)<=0
得sin(x/2+π/3)>=0
所以：0<=x/2+π/3<=π
所以：-π/3<=x/2<=2π/3
解得：-2π/3<=x<=4π/3
所以：单调递减区间为[-2π/3,4π/3]
y=cos(x/2+π/3)
求导：
y'(x)=-sin(x/2+π/3)*(1/2)
=-(1/2)*sin(x/2+π/3)
因为：-2π<=x<=2π
所以：-π<=x/2<=π,-2π/3<=x/2+π/3<=4π/3
所以：-1<=sin(x/2+π/3)<=1
解y'(x)=-(1/2)sin(x/2+π/3)<=0
得sin(x/2+π/3)>=0
所以：0<=x/2+π/3<=π
所以：-π/3<=x/2<=2π/3
解得：-2π/3<=x<=4π/3
所以：单调递减区间为[-2π/3,4π/3]
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