help me solve this transdental equation using matlab ?

bn=((2*vdc)/(n*pi))*{1-2cos*n*a1+2cos*n*a2-2cos*n*a3+2cos*n*a4}.

FOR n=1,5,7,11 And equate b5=b7=b11=0

b1 is known (take any value) and find out a1 a2 a3 a4 ?????

[Walter Roberson]

Where you have 2cos*n*a1 is that intended to mean 2*cos(n*a1) ?

[Adithya Kesineni]

yeah it is 2*cos(n*a1)

Answer 1

[Youssef Khmou]

try, this , Warning : it computationally heavy!!!!

 % This is Non linear system
 % lets take 2*vdc/n*pi=1 for simplification and b1=0 .
 tic,
 Solution=solve('1-2*cos(a1)+2*cos(a2)-2-cos(a3)+2*cos(a4)=2',...
 '1-2*cos(5*a1)+2*cos(5*a2)-2-cos(5*a3)+2*cos(5*a4)=0',...
 '1-2*cos(7*a1)+2*cos(7*a2)-2-cos(7*a3)+2*cos(7*a4)=0',...
 '1-2*cos(11*a1)+2*cos(11*a2)-2-cos(11*a3)+2*cos(11*a4)=0',...
'a1','a2','a3','a4');
 toc,

!!! it will take a long time

[Adithya Kesineni]

Solution =

    a1: [1x1 sym]
    a2: [1x1 sym]
    a3: [1x1 sym]
    a4: [1x1 sym]............this is the output displayed .could u help me  to get  finite values as o/p here.....thnk u

[Youssef Khmou]

hi, well take a look at that solution:

 a1=Solution.a1;
 a2=Solution.a2;
 a3=Solution.a3;
 a4=Solution.a4;

[Walter Roberson]

I don't think you should be getting 1x1 sym.

If you add 2*Pi to any of the a* variables, then when you multiply that addition by the integer n, you are going to be adding integer*2*Pi to the value whose cos() is being taken, and that will have the same result as if the 2*Pi was not there. Therefore there are an infinite number of solutions for each of the a* variables.