F（x)^2=sin^2θcos^4θ=sin^2θsin^2θcos²θ
然后用三元基本不等式求时间来不及打仔细，具体的我等下再打出来F（x)^2=sin^2θcos^4θ=sin^2θcos^2θcos²θ=1/2*2sin^2θcos^2θcos²θ<=[(2sin^2θ+cos^2θ+cos²θ)/3]^3=8/27
所以f(x)<=2根6/9.
当根2sinθ=cosθ时取得等号