函数y=cos2x*cos(2x+π/6)（0≤x≤π/4）的最大值为

y=cos2x(cos2xcosπ/6-sin2xsinπ/6)
=√3/2*(cos2x)^2-1/2sin2xcos2x
=√3/2*(1+cos4x)/2-1/4*sin4x
=-(1/4)(sin4x-√3cos4x)+√3/4
=-(1/4)*2sin(4x-z)+√3/4
其中tanz=-√3
y最大则sin(4x-z)最小

0≤x≤π/4
所以0≤4x≤π
sin4x最小=0
所以y最大=√3/4

y=cos2x*cos(2x+π/6)
=cos2xcos2xcosπ/6-cos2xsin2xsinπ/6
=√3/2*cos^2 2x-1/2cos2xsin2x
=√3/2*(cos4x+1)/2-sin4x/4
=(√3/2*cos4x-1/2*sin4x)/2+√3/4
=sin(π/3-4x)/2+√3/4
最大值为:1/2+√3/4