已知fx=cos二分之πxcos二分之x-sin二分之三xsin二分之x-2sinxcosx,当x属于二分之π,π时,fx的零点为
问题描述：
已知fx=cos二分之πxcos二分之x-sin二分之三xsin二分之x-2sinxcosx,当x属于二分之π,π时,fx的零点为
答
f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx
=cos(3x/2+x/2)-2sinxcosx
=cos2x-sin2x
=√2(√2/2 *cos2x- √2/2 *sin2x)
=√2 sin(π/4-2x)
f(x)=0
sin(π/4-2x)=0
π/4-2x=kπ
x=π/8-kπ/2
当k=-1时,x=5π/8