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

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

f(x)-g(x)=srq(2)/2 *sin(2x+2pi/3-pi/4)=sqr(2)/2 sin(2x+5pi/12) 令2x+5pi/12=Kpi 解得到x=(Kpi-5pi/12)/2

应用辅助角公式！Asinα+Bcosα=(A^2+B^2)sin(ɑ+β) 其中tanβ=B/A
应用该公式上式可以化成h(x)=1/2sin(2x+2π/3+β) β=mπ-π/4 .
则有2x+2π/3+β=nπ
解得
x=｛(n-m)π-5π/12｝/2
即x=｛kπ-5π/12｝/2
哎！我的脑子不够用了啊！