解f(x) = 1/2 x^2 - sinxf'(x) = x - cosxf''(x) = 1+sinx >=0,所以 f'(x)是增函数.f'(0) = -1,f'(2pi) = 2pi -1 >0说明f(x) 从0到2pi先递减再递增f(0) = 0f(2pi) = 2pi^2 >0因此 f(x)在 【0,2pi】上有两个零点,其中...