Question

用MATLAB求下面不定方程的解

β的范围设定为0-pi/2，r=1.4，d=1000，h=1.5；

Answer 1

syms a b d
a=solve('1.4/sqrt(1.5^2-1.4^2)=cos(b)*d*cos(a)/(2*1.5+d*sin(a))','a')
结果：
a =
atan((-3.+10000000000000000000000000000000.*cos(b)*(30000000000000000000000000000000.*cos(b)+(.10000000000000000000000000000000e63*cos(b)^2*d^2-.60827586206896551724137931034485e64+.67586206896551724137931034482761e63*d^2)^(1/2))/(.10000000000000000000000000000000e63*cos(b)^2+.67586206896551724137931034482761e63))/d,25997347344787260840520671338471.*(30000000000000000000000000000000.*cos(b)+(.10000000000000000000000000000000e63*cos(b)^2*d^2-.60827586206896551724137931034485e64+.67586206896551724137931034482761e63*d^2)^(1/2))/(.10000000000000000000000000000000e63*cos(b)^2+.67586206896551724137931034482761e63)/d)
atan((-3.+10000000000000000000000000000000.*cos(b)*(30000000000000000000000000000000.*cos(b)-1.*(.10000000000000000000000000000000e63*cos(b)^2*d^2-.60827586206896551724137931034485e64+.67586206896551724137931034482761e63*d^2)^(1/2))/者茄(.10000000000000000000000000000000e63*cos(b)^2+.67586206896551724137931034482761e63))/d,25997347344787260840520671338471.*(30000000000000000000000000000000.*cos(b)-1.*(.10000000000000000000000000000000e63*cos(b)^2*d^2-.60827586206896551724137931034485e64+.67586206896551724137931034482761e63*d^2)^(1/2))/(.10000000000000000000000000000000e63*cos(b)^2+.67586206896551724137931034482761e63)/d)
>>
syms a b d
b=solve('1.4/sqrt(1.5^2-1.4^2)=cos(b)*d*cos(a)/(2*1.5+d*sin(a))','b')
结果：
b =
acos(2.5997347344787260840520671338471*(3.+d*sin(a))/d/cos(a))
d=solve('饥嫌裤1.4/sqrt(1.5^2-1.4^2)=cos(b)*d*cos(a)/(2*1.5+d*sin(a))','d')
结果：烂简
d =
77992042034361782521562014015413./(-25997347344787260840520671338471.*sin(a)+10000000000000000000000000000000.*cos(b)*cos(a))
>>
说明：a=α,b=β,d=d