设函数f(x)=cos(根号3 *x+y)(0
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设函数f(x)=cos(根号3 *x+y)(0
设函数f(x)=cos(根号3 *x+y)(0
设函数f(x)=cos(根号3 *x+y)(0
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
f(x)=cos[(√3)x+y]
f(x)=cos(x√3+y)
f'(x)=-(√3)sin(x√3+y)
设：g(x)=f(x)+f'(x)
g(x)=cos(x√3+y)-(√3)sin(x√3+y)
g(x)=2[(1/2)cos(x√3+y)-(√3/2)sin(x√3+y)]
g(x)=2[sin(π/6)cos(x√3+y)...
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f(x)=cos[(√3)x+y]
f(x)=cos(x√3+y)
f'(x)=-(√3)sin(x√3+y)
设：g(x)=f(x)+f'(x)
g(x)=cos(x√3+y)-(√3)sin(x√3+y)
g(x)=2[(1/2)cos(x√3+y)-(√3/2)sin(x√3+y)]
g(x)=2[sin(π/6)cos(x√3+y)-cos(π/6)sin(x√3+y)]
g(x)=2sin(π/6-x√3-y)
因为：g(-x)=-g(x)
所以：
2sin[π/6-(-x)√3-y]=-2sin(π/6-x√3-y)
sin[x√3-(y-π/6)]=sin[x√3+(y-π/6)]
sin(x√3)cos(y-π/6)-cos(x√3)sin(y-π/6)=sin(x√3)cos(y-π/6)+cos(x√3)sin(y-π/6)
sin(y-π/6)=0
y-π/6=2kπ
y=2kπ+π/6，其中：k=0、±1、±2、±3、……
已知：0＜y＜π
所以：y=π/6
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