已知函数fx=2cos²x/2-根号3sinx 求函数的最小正周期和值域

f(x)=2cos²(x/2)-√3sinx
f(x)=2cos²(x/2)-2√3sin(x/2)cos(x/2)
f(x)=2cos(x/2)[cos(x/2)-√3sin(x/2)]
f(x)=4cos(x/2)[(1/2)cos(x/2)-(√3/2)sin(x/2)]
f(x)=4cos(x/2)[sin(π/6)cos(x/2)-cos(π/6)sin(x/2)]
f(x)=4cos(x/2)sin(π/6+x/2)
f(x)=2[sin(x+π/6)+sin(π/6)]
f(x)=2[sin(x+π/6)+1/2]
f(x)=1+2sin(x+π/6)
因为：-1≤sin(x+π/6)≤1
所以：-1≤f(x)≤3,
即：所求值域是：f(x)∈[-1,3]
因为：x项的系数是1,
所以：所求最小正周期是：2π/1=2π.