Question

已知函数f(x)=1/2 sin2xsinφ+cos^2xcosφ-1/2sin(π/2+φ)（0＜φ＜π）图像过点(π/6,1/2)

已知函数f(x)=1/2 sin2xsinφ+cos^2xcosφ-1/2sin(π/2+φ)（0＜φ＜π）图像过点(π/6,1/2) 1、求φ的值 2若不等式2f（x+π/6)<cos^2x-2m·sinx+2m+1对x∈[0，π/2]恒成立，求实数m的取值范围

Answer 1

f(x)=1/2 sin2xsinφ+cos²xcosφ-1/2sin(π/2+φ)
=1/2 *2sinxcosxsinφ+cos²xcosφ-1/2cosφ
=sinxcosxsinφ+cos²xcosφ-1/2cosφ
1/2=sinπ/6cosπ/6sinφ+cos²π/6cosφ-1/2cosφ
1/2=1/2*√3/2sinφ+(√3/2)²cosφ-1/2cosφ
1/2=√3/4sinφ+3/4cosφ-1/2cosφ
1=√3/2sinφ+1/2cosφ
1=cosπ/6sinφ+sinπ/6cosφ
sin(φ+π/6)=1
0<φ<π, π/6<φ+π/6<7π/6
φ+π/6=π/2
φ=π/3

f(x)=sinxcosxsinπ/3+cos²xcosπ/3-1/2cosπ/3=√3/2sinxcosx+1/2cos²x-1/4
=√3/4sin2x+(2cos²x-1)/4=√3/4sin2x+cos2x/4=(√3/2sin2x+1/2cos2x)/2
=(sin2xcosπ/6+cos2xsinπ/6)/2=1/2sin(2x+π/6)
2f(x+π/6)=2*1/2sin[2(x+π/6)+π/6]=sin(2x+π/2)=cos2x=2cos²x-1
2cos²x-1<cos²x-2msinx+2m+1
1-cos²x-2msinx+2m+1>0
sin²x-2msinx+2m+1>0
sin²x-2msinx+m²-m²+2m+1>0
(sinx-m)²-m²+2m+1>0
x∈[0，π/2], sinx∈[0，1]
sinx=m∈[0，1]时, (sinx-m)=0
m²-2m-1<0
1-√2<m<1+√2
所以 m∈[0，1]

追问

答案错掉诶.

追答

答案是多少?