y=cos2x+4sinx+1
=1-2sin^2x+4sinx+1
=-2(sin^2x-2sinx-1)
=-2[(sinx-1)^2-2]
=-2(sinx-1)^2+4
sinx=-1时,y取到最小值-4,
sinx=1时,y取到最大值4,
所以y=cos2x+4sinx+1值域为[-4,4]；
y=3cos2x=3sin(2x+π/2)=3sin2(x+π/4),
而y=3sin(2x-π/6)=3sin2(x-π/12)=3sin2[(x+π/4)-π/3],
所以需要往右平移π/3个单位.