数学函数y=sinx+cos2x(0<=x<=2π)
y=sinx+cos2x(0<=x<=2π)的值域是? 答案:[-2,9/8] 请写出过程
y=sinx+cos2x(0<=x<=2π) =sinx+1-2sin^x(0<=x<=2π) =-2(sinx-1/4)^+9/8 0<=x<=2π -1<=sinx<=1 当x=3/2pi时，y最小=-2 当sinx=1/4时，y最大=9/8