函数f（x）=cos2x+sinx在区间[-π4
网友回答
f（x）=cos2x+sinx=1-sin2x+sinx=-（sinx-12
======以下答案可供参考======
供参考答案1:
f=1-snx*sinx+sinx=3/4-(sinx-1/2)*sin(x-1/2)
[-π/4，π/4] -√2/2f(-π/4)=1/2-√2/2 f(π/4)=1/2+√2/2 f(π/6)=3/4
最小值是f(-π/4)=1/2-√2/2
供参考答案2:
答案看图片 函数f（x）=cos2x+sinx在区间[-π4(图1)