急！！高中数学!!帮忙！！谢谢！！

f(x)=2COS X(SIN X+COS X)-1=2cosXsinX+2(cosx)方-1
=sin2x+cos2x=根号2sin(2x+pi/4)
向左平移a个单位就是g(x)=f(x+a)=根号2sin(2(x+a)+pi/4)=根号2sin(2x+2a+pi/4)
g(x)是偶函数,根据sin函数图像特点可以确定,x=0时应取极值
所以 sin(2a+pi/4)=1或-1
0<a<pi/2,所以a=pi/8 或 5pi/8

f(x)=2COS X(SIN X+COS X)-1
=2sinxcosx+2(c0sx)^2-1
=sin2x+cos2x
=√2sin(2x+π/4)
左移a个单位变为：
g(x)=√2sin(2(x+a)+π/4)
=√2sin(2x+2a+π/4)
g(x)是偶函数，则 2a+π/4＝π/2
（此时g(x)变为=√2cos2x,是偶函数）
a=π/8

前同得
g(x)=√2sin(2x+2a+π/4)
g(x)为偶函数
得 g(x)=g(-x)
√2sin(2x+2a+π/4)= √2sin(-2x+2a+π/4)= -√2sin(2x-2a-π/4)
Sin(2x+2a+π/4)+sin(2x-2a-π/4)=0
得 2sin[(2x+2a+π/4+2x-2a-π/4)/2]cos[(2x+2a+π/4-2x+2a+π/4)/2]=0
得 sin2xcos(2a+π/4)=0 (对任意x成立)
cos(2a+π/4)=0
2a+π/4=(2k+1)π/2
所以 2*0+π/4<2a+π/4<2*π/2+π/4=5π/4
期间只有 2a+π/4=π/2
a=π/8