Question

求数学学霸。解题，有详细过程谢谢

Answer 1

f(x)=(1/2)sin2x - √3•[(cos2x + 1)/2]
=(1/2)sin2x - (√3/2)cos2x - √3/2
=sin(2x - π/3) - √3/2
∴T=2π/2=π
最小值是-1 - √3/2

追答

由已知：T=π•2=2π
∴w=2π/2π=1
则g(x)=sin(x - π/3) - √3/2
∵π/2≤x≤π
∴π/6≤x - π/3≤2π/3
则g(x)∈[1/2 - √3/2，1 - √3/2]