运用牛顿二元迭代法在Matlab求解theta1和theta2，u1=(ptheta1+roucos(lambadab)-z)/sin(lambadab) u2=(ptheta2+roucos(lambadab)+z)/sin(lambadab) rbcos(theta1+mu)+u1cos(lambadab)sin(theta1+mu)+rousin(lambadab)sin(theta1+mu)==rbcos(theta2+mu)+u2cos(lambadab)sin(theta2+mu)+rousin(lambadab)sin(theta2+mu); rbcos(theta1+mu)-u1cos(lambadab)cos(theta1+mu)-rousin(lambadab)sin(theta1+mu)==-rbsin(theta2+mu)+u2*cos(lambadab)cos(theta2+mu)+rousin(lambadab)*cos(theta2+mu);

以下是使用牛顿二元迭代法在MATLAB中求解theta1和theta2的代码示例：

% 初始化参数
theta1 = 0;
theta2 = 0;
max_iter = 100;
tolerance = 1e-6;

% 迭代求解
for i = 1:max_iter
    % 计算u1和u2
    u1 = (p*theta1 + rou*cos(lambadab) - z) / sin(lambadab);
    u2 = (p*theta2 + rou*cos(lambadab) + z) / sin(lambadab);
    
    % 计算雅克比矩阵
    J = [rb*cos(theta1+mu)+u1*cos(lambadab)*sin(theta1+mu)+rou*sin(lambadab)*sin(theta1+mu), -rb*cos(theta2+mu)-u2*cos(lambadab)*sin(theta2+mu)-rou*sin(lambadab)*sin(theta2+mu);
         rb*cos(theta1+mu)-u1*cos(lambadab)*cos(theta1+mu)-rou*sin(lambadab)*sin(theta1+mu), -rb*sin(theta2+mu)+u2*cos(lambadab)*cos(theta2+mu)+rou*sin(lambadab)*cos(theta2+mu)];
     
    % 计算函数值
    f = [rb*cos(theta1+mu)+u1*cos(lambadab)*sin(theta1+mu)+rou*sin(lambadab)*sin(theta1+mu) - rb*cos(theta2+mu)-u2*cos(lambadab)*sin(theta2+mu)-rou*sin(lambadab)*sin(theta2+mu);
         rb*cos(theta1+mu)-u1*cos(lambadab)*cos(theta1+mu)-rou*sin(lambadab)*sin(theta1+mu) + rb*sin(theta2+mu)-u2*cos(lambadab)*cos(theta2+mu)+rou*sin(lambadab)*cos(theta2+mu)];
     
    % 计算更新量
    delta = -J\f;
    
    % 更新theta1和theta2
    theta1 = theta1 + delta(1);
    theta2 = theta2 + delta(2);
    
    % 判断是否收敛
    if norm(delta) < tolerance
        break;
    end
end

% 输出结果
fprintf('theta1 = %f\n', theta1);
fprintf('theta2 = %f\n', theta2);

需要注意的是，以上代码中的参数p、rou、z、lambadab、mu和rb应该在代码中定义并进行赋值。另外，这里的迭代次数max_iter和误差容限tolerance也可以根据实际情况进行调整。