Question

紧急!如何用MATLAB解这个超越方程组

紧急！如何用MATLAB解这个超越方程组
COS(X1)+COS(X2)+COS(X3）=3π/5;
COS(5X1）+COS(5X2)+COS(5X3）=0；
COS(7X1）+COS(7X2)+COS(7X3）=0
x为角度,范围是0-π/2

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Answer 1

syms x1 x2 x3 real
assume(0<x1<pi/2);
assume(0<x2<pi/2);
assume(0<x3<pi/2);
eq1='cos(x1)+cos(x2)+cos(x3)=3/5*pi';
eq2='cos(5*(x1))+cos(5*(x2))+cos(5*(x3))=0';
eq3='cos(7*(x1))+cos(7*(x2))+cos(7*(x3))=0';
[x1,x2,x3]=solve(eq1,eq2,eq3,'x1','x2','x3')
运行结果：
x1 =
1.1254646268190320378701736260237

x2 =
0.5102556983259647809687583398787

x3 =
0.95012834533048942017160732079172