how to solve the below non linear algebraic equation using matlab

How do we solve these two equations?

1.5 cos60 + 4.0 cos(x3) - 3.5 cos(x4) - 3.0 = 0

1.5 sin60 + 4.0 sin(x3) - 3.5 sin(x4) = 0

Since these are nonlinear algebraic equations, they can be solved numerically by iterative methods such as Newton-Raphson

Answer 1

[Azzi Abdelmalek]

save this function myeq

function f=myeq(x)
f=[1.5*cos(60) + 4.0*cos(x(1)) - 3.5* cos(x(2)) - 3.0 
  1.5 *sin(60) + 4.0*sin(x(1)) - 3.5*sin(x(2))]

then call it

 sol=fsolve(@myeq,[0;0])

Answer 2

[Roger Stafford]

You can perhaps get more insight into the solution of your equations if you proceed along lines such as the following. For convenience in notation rewrite the equations as

 a*cos(u) = b*cos(v) + c
 a*sin(u) = b*sin(v) + d

where u = x3, v = x4, a = 4, b = 3.5, c = 3-1.5*cos60, d = -1.5*sin60. Now square both sides of both equations and add them to obtain

 a^2*cos(u)^2+a^2*sin(u)^2 = (b*cos(v)+c)^2+(b*sin(v)+d)^2
 a^2 = b^2+2*b*(c*cos(v)+d*sin(v))+c^2+d^2
 c*cos(v)+d*sin(v) = (a^2-b^2-c^2-d^2)/(2*b)

Now divide both sides by sqrt(c^2+d^2)

 (c/sqrt(c^2+d^2))*cos(v)+(d/sqrt(c^2+d^2))*sin(v) = 
    (a^2-b^2-c^2-d^2)/(2*b*sqrt(c^2+d^2))

Use atan2 to get

 t = atan2(d,c)

which will satisfy c = sqrt(c^2+d^2)*cos(t), d = sqrt(c^2+d^2)*sin(t) so that we get

 cos(t)*cos(v)+sin(t)*sin(v) = cos(v-t) =
    (a^2-b^2-c^2-d^2)/(2*b*sqrt(c^2+d^2))

 v = t + acos((a^2-b^2-c^2-d^2)/(2*b*sqrt(c^2+d^2)))

This will have two solutions within the interval from -pi to +pi. Then to get u from v do

 u = atan2((b*sin(v)+d)/a,(b*cos(v)+c)/a)

In other words it all reduces to the two matlab lines:

 v = atan2(d,c) + acos((a^2-b^2-c^2-d^2)/(2*b*sqrt(c^2+d^2)));
 u = atan2((b*sin(v)+d)/a,(b*cos(v)+c)/a);

except that you have to allow acos to have two solutions, one the negative of the other.

Much more satisfying in my opinion than a mindless solution from 'fsolve' where you may not be informed that there are infinitely many solutions (though of course more mental effort.)

Roger Stafford

[P Anand Raju]

hi i tried the last equation i.e.

v = atan2(d,c) + acos((a^2-b^2-c^2-d^2)/(2*b*sqrt(c^2+d^2)))

u = atan2((b*sin(v)+d)/a,(b*cos(v)+c)/a)

and used another equation to just check the error and i am getting some values which should be 0 or near to 0 .

e1=a*cos(u) - b*cos(v) - c

e2=a*sin(u)- b*sin(v) - d

[Roger Stafford]

I assumed by sin60 and cos60 you meant the sine and cosine of 60 degrees. I used those formulas I gave you with these values and arrived at

 e1 = 0
 e2 = 2.2204e-16

For the other pair with the acos result sign reversed I got

 e1 = -4.4409e-16
 e2 = -4.4409e-16

That's about as close to zero as you are going to get on a machine with 53-bit significands.

Roger Stafford