已知函数f(x)=(1/2)cos(2x+2π/3),g(x)=(1/2)sin(2x+2π/3).
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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.
求函数h(x)=f(x)-g(x)的零点.
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+π/2,k∈Z.
X= kπ/2-5π/24,k∈Z.
这就是函数h(x)=f(x)-g(x)的零点.
=√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+π/2,k∈Z.
X= kπ/2-5π/24,k∈Z.
这就是函数h(x)=f(x)-g(x)的零点.
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