k=tana
所以-√3<=tana<=√3/3
若-√3=tan2π/3<=tana<0
因为0<=a<π
所以2π/3<=a<π
若0<=tana<=√3/3=tanπ/6
则0<=a<=π/6
所以[0,π/6]∪[2π/3,π)