设函数f(x)=cos(√3x+θ) (0
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设函数f(x)=cos(√3x+θ) (0
![设函数f(x)=cos(√3x+θ) (0](/uploads/image/z/6108664-40-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%28%E2%88%9A3x%2B%CE%B8%29+%280)
f(x)+f'(x)=cos(√3x+θ)-√3sin(√3x+θ)
f(0)+f'(0)=cos(θ)-√3sin(θ)=0
tan(θ)=√3
θ=pi/3
f(0)+f'(0)=cos(θ)-√3sin(θ)=0
tan(θ)=√3
θ=pi/3
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