In the first half, sinα= 1×sinα, and the coefficients before the two trigonometric functions are 1 and √3, so we will think of a special angle. With the auxiliary angle formula, the first half can be reduced, and α has a range, so the first half has a range, so adding 1 is the range of m.

2 (1/2sinα-√ 3/2cosα) = 2 (sinα cos60-sin60cosα) = 2sin (α-60), because α belongs to (0, π).

So α-60 ∈ (-60， 120)，so 2 sin (α-60) ∈ (-√ 3，√ 3)，so-√ 3 < m- 1 < √ 3，

So 1-√ 3 < m < 1+√ 3.