已知函数F[x]=sinxcosx+cos^2x-1/2,求最小正周期.若f[x]在区间[0,π/2]上的最大值和最小值及相应的x值

F[x]=sinxcosx+cos^2x-1/2
=1/2sin2x+1/2(cos2x+1)-1/2
=1/2(sin2x+cos2x)
=√2/2sin(2x+π/4)
最小正周期T=2π/W=π
2x+π/4=2Kπ+π/2,X=kπ+π/8
x=π/8,x=π+π/8 ,f(x)最大值=√2/2
2x+π/4=2Kπ-π/2 ,x=kπ-3π/8
x=5π/8,x=13π/8 ,f(x)最小值=-√2/2