那么x的取值只有一點(diǎn)-2π/3,y應(yīng)該只有一個(gè)值,那么值域怎么可能是【-1/4,2】呢
∴a不能取到-2π/3
y=sin^2x+2cosx
=1-cos^2x+2cosx
=-(cos^2x-1)+2
令-(cos^2x-1)+2=-1/4
cosx=5/2(舍去)或-1/2
∴cosx=-1/2,即
x=2π/3+2kπ或-2π/3+2kπ,k∈Z
令-(cos^2x-1)+2=2
cosx=1
x=2kπ,k∈Z
![](http://d.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=541e984a8326cffc697fb7b4893166a8/7e3e6709c93d70cfd89c44def9dcd100bba12bfb.jpg)
x∈[-2π/3,a]
根據(jù)圖像
a的范圍(0,2π/3]【-2π/3,2π/3】因?yàn)閍=-2π/3有最小值-1/4能取到