求解f(x)=2cosX^2+3sinX+3 的值域 X属于π/6到2π/3

f(x)=2cos²x+3sinx+3
=2(1-sin²x)+3sinx+3
=-2sin²x+3sinx+5
=-2(sin²x-3sinx/2)+5
=-2(sin²x-3sinx/2+9/16)+5+9/8
=-2(sinx -3/4)²+49/8
x∈[π/6,2π/3]
1/2≤sinx≤1
当sinx=3/4时,f(x)有最大值[f(x)]max=49/8
当sinx=1/2时,f(x)有最小值[f(x)]min=-2(-1/2-3/4)²+49/8=3
函数的值域为[3,49/8].