原式可化为 f(x)=根3sinx-cosx即f(x)=2sin(x-6/π)再求下递增区间0π/6
f(x)=2sin(x+π/6)-2 cosx将式子展开后再合并，可得f(x)=2sin(x-π/6）则在[0,π]区间单调递减区间为π/2< x-π/6<π解得 2π/3< =x<=7π/6
