解f(x)=(2cos²-1)sin2x+1/2cos4x=cos2xsin2x+1/2cos4x=1/2sin4x+1/2cos4x=√2/2sin(4x+π/4)∴T=2π/4=π/2为其最小周期最大值为：√2/2f(a)=√2/2sin(4a+π/4)=√2/2∴4a+π/4=π/2+2kπ∴a=π/16+kπ/2∵a∈(...