f(x)=cos(x+π/6）cos(x-2π/3)+sin(x+π/6)sin(x+π/3) 求函数F（X）的单调递减区间

sin(x+π/3)=-sin(x+π/3-π)=-sin(x-2π/3) f(x)=cos(x+π/6)cos(x-2π/3)+sin(x+π/6)sin(x+π/3)=cos(x+π/6)cos(x-2π/3)-sin(x+π/6)sin(x-2π/3)=cos[(x+π/6)+(x-2π/3)]=cos(2x-π/2)cosx单调递减区间为[2k...