作业帮求非线性方程组的“解析解”
来源：学生作业帮 编辑：作业帮 分类：综合作业 时间：2024/05/27 16:01:54
求非线性方程组的“解析解”
-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0
-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0
-x0*cos(b)-y0*sin(a)*cos(b)-z0*cos(a)*cos(b)=sqrt(x0^2+y0^2+z0^2)
其中x0,y0,z0已知,但不是某个具体的常数.所以就解出这个方程的“解析解”,a,b,c用x0,y0,z0的反三角函数表示.用matlab或mathmatics具体如何编程,希望给出可以运行的源代码.
-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0
-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0
-x0*cos(b)-y0*sin(a)*cos(b)-z0*cos(a)*cos(b)=sqrt(x0^2+y0^2+z0^2)
其中x0,y0,z0已知,但不是某个具体的常数.所以就解出这个方程的“解析解”,a,b,c用x0,y0,z0的反三角函数表示.用matlab或mathmatics具体如何编程,希望给出可以运行的源代码.
程序很简单
[a,b,c]=solve('-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0','-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0','-x0*cos(b)-y0*sin(a)*cos(b)-z0*cos(a)*cos(b)=sqrt(x0^2+y0^2+z0^2)','a','b','c');
但是matlab没有解出解析解.
[a,b,c]=solve('-x0*cos(b)*cos(c)-y0*(-sin(a)*cos(b)*cos(c)-cos(a)*sin(c))-z0*(-cos(a)*cos(b)*cos(c)+sin(a)*sin(c))=0','-x0*cos(b)*sin(c)-y0*(-sin(a)*cos(b)*sin(c)+cos(a)*cos(c))-z0*(-cos(a)*cos(b)*sin(c)-sin(a)*cos(c))=0','-x0*cos(b)-y0*sin(a)*cos(b)-z0*cos(a)*cos(b)=sqrt(x0^2+y0^2+z0^2)','a','b','c');
但是matlab没有解出解析解.