已知函数f(x)=(1/2)cos(2x+2π/3),g(x)=(1/2)sin(2x+2π/3).
问题描述:
已知函数f(x)=(1/2)cos(2x+2π/3),g(x)=(1/2)sin(2x+2π/3).
求函数h(x)=f(x)-g(x)的零点.
答
h(x)=f(x)-g(x)= (1/2)cos(2x+2π/3)- (1/2)sin(2x+2π/3)=√2/2[√2/2 cos(2x+2π/3) -√2/2 sin(2x+2π/3)]=√2/2 cos[(2x+2π/3)+π/4] =√2/2 cos(2x+11π/12).h(x)=0时,cos(2x+11π/12) =0,2x+11π/12=kπ+...