求函数f(x)=1/2sin^2x+cos^2+根号3/4sin2x的周期和值域

f(x)=(1-cos2x)/4+(1+cos2x)/2+√3sin2x/4
=(1-cos2x+2+2cos2x)/4+√3sin2x/4
=(cos2x+3)/4+√3sin2x/4
=(cos2x+sin2x)/4+3/4
=2[cos2x*(1//2)+sin2x*(√3/2)]+3/4
=（1/2）sin(2x+π/6)+3/4,
故其周期T=2π/2=π,
值域f(x)∈[1/4,5/4].