已知cos^2θ+(a-6)sinθ+2a-12≤0对θ∈R恒成立,求a的取值范围

已知cos^2θ+(a-6)sinθ+2a-12≤0对θ∈R恒成立
则cos^2θ+asinθ-6sinθ+2a-12≤0
则a(sinθ+2)≤12+6sinθ-cos^2θ
由于sinθ+2>0
所以a≤(12+6sinθ-cos^2θ)/(sinθ+2)=(11+6sinθ+sin^2θ)/(sinθ+2)
所以只需求(11+6sinθ+sin^2θ)/(sinθ+2)的最小值即可
令t=sinθ,g=(11+6sinθ+sin^2θ)/(sinθ+2)
则g=(11+6t+t^2)/(t+2)=(t+2)+3/(t+2)+2>=2(1+根号3)
所以a

cos^2θ+(a-6)sinθ+2a-12≤0，
a(sinθ+2)≤12+6sinθ-cos²θ
=12+6sinθ-1+sin²θ
=11+6sinθ+sin²θ
=(sinθ+2)²+2(sinθ+2)+3，
即a≤(sinθ+2)+2+ 3/(sinθ+2)对θ∈R恒成立，
令t=sinθ+...

cos²θ=1-sin²θ
设sin²θ=t²
sinθ∈[-1,1]
你肯定能写出来