Question

求此题解法 过程写出来

Answer 1

f(x)=√3sin2x + cos2x
=2[(√3/2)sin2x + (1/2)cos2x]
=2sin(2x + π/6)
∴T=2π/2=π
∵0≤x≤π/2
∴0≤2x≤π
则π/6≤2x + π/6≤7π/6
∴f(x)的最大值是2，f(x)的最小值是-1

追答

f(x0)=2sin(2x0 + π/6)=6/5
∴sin(2x0 + π/6)=3/5
则cos(2x0 + π/6)=±√1 - sin²(2x0 + π/6)=±4/5
∵π/4≤x0≤π/2
∴π/2≤2x0≤π
则2π/3≤2x0 + π/6≤7π/6
∴cos(2x0 + π/6)=-4/5
则cos2x0=cos[(2x0 + π/6) - π/6]
=cos(2x0 + π/6)cos(π/6) + sin(2x0 + π/6)sin(π/6)
=(-4/5)•(√3/2) + (3/5)•(1/2)
=(3 - 4√3)/10