关于x的方程sinx-根号3cosx=4m-6/4-m有解,则m的取值范围为

3cosx=4m-6/4-m
因为-1≤cosx≤1
得
-3≤(4m-6)/(4-m)≤3
(4m-6)/(4-m)≥-3
(4m-6+12-3m)/(4-m)≥0
(m+6)(m-4)≤0
-6≤m题目看错了吧？ 前面有sinx的啊晕白费了sinx-根号3cosx=4m-6/4-m2sin(x-π/3)=(4m-6)/(4-m)-2≤(4m-6)/(4-m)≤2(4m-6)/(4-m)≥-2(4m-6+8-2m)(m-4)≤0(2m+2)(m-4)≤0-1≤m<4(4m-6)/(4-m)≤2(4m-6-8+2m)(m-4)≥0(6m-14)(m-4)≥0m≤7/3 或 m>4取交集得-1≤m≤7/3