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=-7/16sin2x+tcos2x=√(1+t²)sin(2x+φ)因为x∈...
第二题:sin2x+tcos2x=√(1+t²)sin(2x+φ)因为x∈(π/12,π/6]所以2x∈(π/6,π/3]要使sin2x+tcos2x≥0在区间(π/12,π/6]上恒成立必须使sin(2x+φ)≥0在区间(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]内的值域是[√3/2,1]所以有√3/2≤1/√(1+t²)≤1解得0≤t≤√3/3或者-√3/3≤t≤0