How to solve a equation with a single variable for a range of values?

I have a linear equation with 2 variables. I want to calculate the value of one variable for a given rane of my second variable.

These are my formula and alpha and L2 are my variable. I want to calculate the value of L2 for alpha from 0 to 42 in steps of 1. But when I solve the equation, I still get the answer in terms of L2 and not a numerical value.

Here is my code

clc;
clear all;

%Getting the input of Constant Values
syms L2;
L1=input('Enter the value of L1:');
R1=input('Enter the value of R1:');
R2=input('Enter the value of R2:');
R3=input('Enter the value of R3:');
a=input('Enter the value of a:');

%Rewriting all variables in terms of L2
alpha = 1:1:42;
L3=sqrt(L1^2+L2^2+2*L1*L2.*cos(alpha));
cosbeta = (L1^2+L3.^2-L2^2)./(2*L1.*L3);
beta=acos(cosbeta);
theta1=(180-beta)*(pi/180);
theta2=alpha*(pi/180);
theta3=(180-alpha+beta)*(pi/180);

%Computing L2
eqn=a-(L1+theta1*R1+theta2*R2+theta3*R3+L3);
solve(eqn,L2);

[David Hill]

You need beta to solve for L2 given an array of alphas.

[Abinav Shankar]

Actually my beta is dependent on L1, L2 and L3. L1 is a constant, L3 is again dependent on L1, L2 and alpha. So technically my beta is dependent on L2 and alpha. I can rewrite my entire equation in terms of L2 and alpha.

Answer 1

[darova]

I think it can be kind of complicated for symbolic calculations

clc,clear
syms a L2
    % define your constants
eqn = % your long expression
ezplot(eqn)

Or another aproach: examine if there are roots (F == 0)

F = matlabFunction( eqn,'vars',[a L2] );
[X,Y] = meshgrid(0:10);
surf(X,Y,F(X,Y))
xlabel('a axis')
ylabel('L2 axis')

[Abinav Shankar]

I think its a bit too complicated. I was able to easily solve it for a single value instead of a range. Then I used the solution in a for loop to generate solution for a range of values. It's slow but atleast it gets the job done.

Thanks.