求函数y=3sin^2x-4cosx+1,x属于[π/3,2π/3]的值域_数学解答

求函数y=3sin^2x-4cosx+1,x属于[π/3,2π/3]的值域

人气：499 ℃　时间：2019-12-19 09:27:33

解答

y=3sin^2x-4cosx+1
y=3(1-cos^2x)-4cosx+1
=3-3cos^2x-4cosx+1
=-3cos^2x-4cosx+4
=-3(cosx-2/3)^2+16/3.
因为x属于[π/3,2π/3],
所以,-1/2