Решение нелинейного уравнения методом секущих

#include "stdafx.h"
#include <iostream>
#include <cmath>
 
int N = 0;
 
double f(double x)
{
    return 2 * sin(x) - x + 0.4;
}
 
double Metod_hord(double a, double b, double eps)
{
    while (fabs(b - a) > eps)
    {
        a = b - (b - a) * f(b) / (f(b) - f(a));
        b = a + (a - b) * f(a) / (f(a) - f(b));
        N++;
    }
    return b;
}
 
int main()
{
    double eps = 0.01;
    double x_begin = -5.;
    double x_end = 0.;
    double x_sr = x_begin;
    double F;
 
    do {
        x_sr += 1.;
        F = f(x_sr);
    } while (F > 0);
 
    double X1 = Metod_hord (x_begin, x_sr,eps);
 
    std::cout << " X1 = " << X1 << "   " << " N = " << N << std::endl;
    
    N = 0;
    double X2 = Metod_hord(x_sr, x_end, eps);
 
    std::cout << " X2 = " << X2 << "   " << " N = " << N << std::endl;
    
    system("pause");
}

#include "stdafx.h"
#include <iostream>
#include <cmath>
 
double f(double x)
{
    return 2. * sin(x) - x + 0.4;
}
 
double Metod_s(double x1, double x0, double eps)
{
    double  x_next;
    int     N = 0;
    while (fabs(x1 - x0) > eps)
    {
        x_next = x0 - f(x0) * (x1 - x0) / (f(x1) - f(x0));
        x0 = x1;
        x1 = x_next;
        N++;
    }
    std::cout << " N = " << N << "   ";
    return x1;
}
 
int main()
{
    double eps = 0.01;
    double x_begin = -5.;
    double x_end = 0.;
 
    double X1 = Metod_s(-3., x_begin, eps);
 
    std::cout << " X1 = " << X1 <<std::endl;
 
    double X2 = Metod_s(x_end, 1., eps);
 
    std::cout << " X2 = " << X2 << std::endl;
 
    system("pause");
}