f(x)=4cos²2x-8cos2x+3=4(cos2x-1)²-1>0(cos2x-1)²>1/4cos2x-1<-1/2,cos2x-1>1/2-1<=cos2x<1/2,cos2x>3/2不成立-1<=cos2x<1/2=cos(2kπ±π/3)所以2kπ-π/2<2x<2kπ+π/2所以kπ-π/4
把COS2X整个当成X解噻，限定个范围就行了
