因为f(x)+f'(x)为奇函数,所以设P=f(x)+f'(x)=cos(√3 *x+y)-√3sin(√3 *x+y)=2(1/2cos(√3 *x+y))-(√3 /2sin(√3 *x+y))=2cos(√3 *x+y+π/3)令P=0,所以cos(y+π/3)=0所以y+π/3=π/2+2kπ所以y=π/6+2kπ又因为0...