解:由題意知,奇函數(shù)f(x)在R上是增函數(shù),f(cos2θ-3)+f(4m-2mcosθ)>f(0)可f(cos2θ-3)>f(-4m+2mcosθ),即cos2θ-3>-4m+2mcosθ,
即cos2θ-3>m(2cosθ-4),由于2cosθ-4<0,故得m>

=

=4+cosθ-2+

,由于4+cosθ-2+

≤4-2

,所以m>4-2

即存在m>4-2

使f(cos2θ-3)+f(4m-2mcosθ)>f(0)對[0,

]恒成立,
答:存在存在m∈R,使f(cos2θ-3)+f(4m-2mcosθ)>f(0)對[0,

]恒成立,m的范圍是m>4-2

分析:奇函數(shù)f(x)定義域R,故f(0)=0,不等式f(cos2θ-3)+f(4m-2mcosθ)>f(0)可轉(zhuǎn)化為f(cos2θ-3)>f(-4m+2mcosθ),再由f(x)在[0,+∞)為增函數(shù),得在R上是增函數(shù),由單調(diào)性解不等式即可.
點評:本題考查奇偶性與單調(diào)性的綜合,綜合考查了利用函數(shù)的性質(zhì)解抽象不等式恒成立的問題,本題綜合性較強(qiáng),比較抽象,解決本題的關(guān)鍵是靈活運(yùn)用函數(shù)的性質(zhì)進(jìn)行正確轉(zhuǎn)化.