Let g (x) = 2sin2x-2msinx+m 2+m-1.

g'(x)=2cosx(2sinx-m)

Jean 2 sinx-m

G(x) increases monotonically. Namely: g (pi) = m2+m-1>; 0

Find m; (Root number 5- 1)/2

And because m & gt=0, m> (radical number 5- 1)/2.

Similarly, set 2sinx-m >; 0，m

Similarly, when 2sinx-m=0, find the second derivative = 4cos2x >; =0, and the points equal to 0 are finite, and the first derivative is 0, then when m=sinx, the minimum value of the function is-1

To sum up: m> (radical number 5- 1)/2 or m.