sinx-√3cosx=√[1^2+(√3)^2]*sin[x+arctan(-√3)]
=2*sin(x-π/3)
所以左边取值范围是 大于等于-2,小于等于2
所以 -2<=(4a-6)/(4-a)<=2
-2<=(4a-6)/(4-a)
(2a-3)/(4-a)+1>=0
(2a-3+4-a)/(4-a)>=0
(a+1)(4-a)>=0
(a+1)(a-4)<=0
-1<=a<=4
(4a-6)/(4-a)<=2
(2a-3)/(4-a)-1<=0
(2a-3-4+a)/(4-a)<=0
(3a-7)(4-a)<=0
(3a-7)(a-4)>=0
a<=7/3,a>=4
综上,再考虑4-a在分母,a不等于4
所以-1<=a<=7/3