设2f(x)cos x=d/dx [f(x)]²,f(0)=1,则f(x)=

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设2f(x)cosx=d/dx[f(x)]²,f(0)=1,则f(x)=设2f(x)cosx=d/dx[f(x)]²,f(0)=1,则f(x)=设2f(x)cosx=d/dx[f(

设2f(x)cos x=d/dx [f(x)]²,f(0)=1,则f(x)=
设2f(x)cos x=d/dx [f(x)]²,f(0)=1,则f(x)=

设2f(x)cos x=d/dx [f(x)]²,f(0)=1,则f(x)=
因为2f(x)cos x=d/dx [f(x)]²=2f(x)f'(x),所以2f(x)[f'(x)-cosx]=0,有f'(x)=cosx
得:f(x)=sinx+C
因为f(0)=1,所以f(x)=sinx+1