f(x)=cos^2x+2sinxcosx-sin^2x=cos^2x-sin^2x+2sinxcosx=cos2x+sin2xf(α/2)=cosα+sinα=3/4（cosα+sinα）²=（3/4）²1+sin2α=9/16sin2α=9/16-1=-7/16 sin2x+tcos2x=√(1+t²)sin（2x+φ）因為x∈...謝謝，那第二題呢第二題：sin2x+tcos2x=√(1+t²)sin（2x+φ）因為x∈(π/12,π/6] 所以2x∈(π/6，π/3] 要使sin2x+tcos2x≥0在區(qū)間(π/12,π/6]上恒成立必須使sin（2x+φ）≥0在區(qū)間(π/12,π/6]上恒成立即需要π/6+φ≥0且π/3+φ≤π即-π/6≤φ≤2π/3又cosφ=1/√(1+t²)，且cosφ在[-π/6,2π/3]內的值域是[√3/2,1]所以有√3/2≤1/√(1+t²)≤1解得0≤t≤√3/3或者 -√3/3≤t≤0