已知函数f（x）=2sin²（π/4+x）-根号3cos2x

f(x) = 2sin²(π/4 +x) - √3 cos2x
= 1 - cos(π/2 + 2x) - √3 cos2x
= 1 + sin2x - √3 cos2x
= 1 + 2sin(2x - π/3)
π/4 ≤ x ≤ π/2
π/6 ≤ 2x - π/3 ≤ 2π/3
1/2 ≤ sin(2x - π/3) ≤ 1
2 ≤ f(x) ≤ 3
f(x) - m = 2 即 m+2 = f(x) 有解
只需 2 ≤ m+2 ≤ 3
即 0 ≤ m ≤ 1