f(x)=sin(2x+π/3)+sin(2x-π/3),g(x)=（√3）cos2x
一动点直线x=t与函数y=f(x),y=g(x)的图像分别交于M、N两点
f(x)-g(x)
=sin(2x+π/3)+sin(2x-π/3)-√3cos2x
=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3 -√3cos2x
=2sin2x*1/2-√3cos2x
=sin2x-√3cos2x
=2(1/2sin2x-√3/2cos2x)
=2sin(2x-π/3)
∴|MN|=2|sin(2t-π/3)|
2t-π/3=π/2+2kπ,k∈Z时, |MN|取得最大值为2f(x)-g(x)，为什么这样做？这是什么意思？x=t时，是两个函数横坐标相等，|MN|就是两个y值的差的绝对值呀