f(x)≤|f(π/6)|,
∴sin(π/3+φ)=土1,
π/3+φ=（k+1/2)π,k∈Z.
φ=(k+1/6)π,
由f(π/2)>f(π)得
sin[(k+7/6)π]-sin[(k+13/6)π]>0,
-2sin(π/2)cos[(k+5/3)π>0,取k=-1,
f(x)=sin(2x-5π/6),单调递增区间由
（2k-1/2)π