用matlab 中solve求解线性方程组,求大神帮帮,谢谢啦。。。
symss3,x3,x4,se;[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0...
syms s3,x3,x4,se;
[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0.125*sin(0.174)','0.6*cos(x3)+0.15*cos(x4)-se=0','0.6*sin(x3)+0.15*sin(x4)=0.3')
结果出现:
s3 =
z2
z2
x3 =
(3*cos(z3))/5 + (3*cos(z4))/20
(3*cos(z3))/5 + (3*cos(z4))/20
x4 =
z3
z3
se =
z4
为何呢?应该为数字吧。。。 展开
[s3,x3,x4,se]=solve('s3*cos(x3)=0.125*cos(0.174)','s3*sin(x3)=0.275+0.125*sin(0.174)','0.6*cos(x3)+0.15*cos(x4)-se=0','0.6*sin(x3)+0.15*sin(x4)=0.3')
结果出现:
s3 =
z2
z2
x3 =
(3*cos(z3))/5 + (3*cos(z4))/20
(3*cos(z3))/5 + (3*cos(z4))/20
x4 =
z3
z3
se =
z4
为何呢?应该为数字吧。。。 展开
展开全部
按你的语句 运行完的结果和你说的不一样:
s3 =
-.32117320644214174679478487023195
.32117320644214174679478487023195
x3 =
-.22999275816638841028516780789997+.15000000000000000000000000000000*(-3.-16.*sin(x3)^2+16.*sin(x3))^(1/2)
.22999275816638841028516780789997+.15000000000000000000000000000000*(-3.-16.*sin(x3)^2+16.*sin(x3))^(1/2)
x4 =
-1.9641858908552267692932569383891
1.1774067627345664691693864448904
se =
-1.*asin(4.*sin(x3)-2.)
-1.*asin(4.*sin(x3)-2.)
第一条语句写不写都无所谓吧,本来用solve解出来的就是符号解,就不需要再把s3,x3,x4,se定义成符号,而且syms s3,x3,x4,se;的逗号应该换成空格。
s3 =
-.32117320644214174679478487023195
.32117320644214174679478487023195
x3 =
-.22999275816638841028516780789997+.15000000000000000000000000000000*(-3.-16.*sin(x3)^2+16.*sin(x3))^(1/2)
.22999275816638841028516780789997+.15000000000000000000000000000000*(-3.-16.*sin(x3)^2+16.*sin(x3))^(1/2)
x4 =
-1.9641858908552267692932569383891
1.1774067627345664691693864448904
se =
-1.*asin(4.*sin(x3)-2.)
-1.*asin(4.*sin(x3)-2.)
第一条语句写不写都无所谓吧,本来用solve解出来的就是符号解,就不需要再把s3,x3,x4,se定义成符号,而且syms s3,x3,x4,se;的逗号应该换成空格。
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