1. Solution: (1) As can be seen from the diagram in the problem, T2=5π6-π3, that is, T=π, so ω=2πT=2.

According to the five-point drawing method, let 2×π3+φ=π, and φ = π 3 is obtained.

So f (x) = cos (2x+π 3);

(2) combining (1) to make an image of the function f(x)=cos(2x+π3) on [-π 6, 13 π 12],

It can be seen from the image that when m= 1 or m ∈ (- 1, 0), there are two different real roots.

2.√3

3, 1\2

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