f(x)=-[sin(3x-i)-√3cos(3x-i)]
=-2sin(3x-i-π/3)
是奇函數(shù)
則q(x)=sin(3x-i-π/3)是奇函數(shù)
g(x)+g(-x)=0
sin(3x-i-π/3)+sin(-3x-i-π/3)=0
sin(3x-i-π/3)=-sin(-3x-i-π/3)=sin(3x+i+π/3)
所以3x-i-π/3=-2kπ+3x+i+π/3或3x-i-π/3=-2kπ+π+3x+i+π/3
-i-π/3=-2kπ+i+π/3
2i=2kπ-2π/3
i=kπ-π/3
3x-i-π/3=-2kπ+π+3x+i+π/3
2i=2kπ-5π/3
i=kπ-5π/6
所以i=kπ-π/3或i=kπ-5π/6