问题标题：
已知函数f(x)=1/2cos(2x+2π/3),g(x)=1/2sin(2x+2π/3).求函数h(x)=f(x)-g(x)的零点
问题描述：
已知函数f(x)=1/2cos(2x+2π/3),g(x)=1/2sin(2x+2π/3).求函数h(x)=f(x)-g(x)的零点
金荆回答：
h(x)=f(x)-g(x)
=1/2cos(2x+2π/3)-1/2sin(2x+2π/3)
=√2/2*[√2/2cos(2x+2π/3)-√2/2sin(2x+2π/3)]
=√2/2*[cos(2x+2π/3)cosπ/4-sin(2x+2π/3)sinπ/4]
=√2/2*cos(2x+2π/3+π/4)
=√2/2*cos(2x+11π/12)
h(x)=0
√2/2*cos(2x+11π/12)=0
cos(2x+11π/12)=0
2x+11π/12=kπ+π/2
2x=kπ-5π/12
x=kπ/2-5π/24
函数h(x)的零点是x=kπ/2-5π/24
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