f(x)=x+2cosx
f'(x)=1-2sinx
令f'(x)=0得sinx=1/2
∵x∈[0,π/2]
∴x=π/6
x∈[0,π/6),f'(x)>0 ,f(x)递增
x∈(π/6,π/2],f'(x)