已知函数f(x)=2sinxcosx+sin(2x+派除以2),若x属于R,求f(x)的最小正周期和单调递增区间
解析：∵函数f(x)=2sinxcosx+sin(2x+π/2)=sin2x+cos2x=√2sin(2x+π/4),
∴f(x)的最小正周期为π
2kπ-π/2