令 f(x)=x-(1/2)sinx,
则f'(x)=1-(1/2)cosx≥1-1/2=1/2>0
从而 f(x)在R上是单调增函数,
又f(0)=0-(1/2)sin0=0,
从而方程x-(1/2)sinx=0有唯一解为x=0