设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为

设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为
设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为

设函数f(x)=cos(2x+π/3)+sin^2x-1/2,当x∈[0,π]时,f(x)的值域为
f(x)=cos(2x+π/3)+sin^2x-1/2=cos(2x+π/3)-1/2cos2x=-√3/2sin2x
当x∈[0,π]时,f(x)的值域为[-√3/2,√3/2]

(x)=cos(2x+π/3)+sin^2x-1/2
=cos2xcosπ/3-sin2xsinπ/3+(1-cos2x)/2-1/2
=1/2cos2x-√3/2*sin2x+1/2-1/2cos2x-1/2
=-√3/2sin2x
x∈[0,π]
2x∈[0,2π]
f(x)的递减区间:(0,π/2)U(3π/2,2π