I need the values of r for different t values, this is how I tried to do it:
syms r
syms t
eq = r^2 == r^2*cos(t)^2 + ((283*r^2*cos(t)^2)/50 + (3481*r*cos(t))/2500)^2;
thetas = [0: pi/20: pi/2];
eq1 = subs(eq, t, thetas);
R = solve(eq1, r)

 Accepted Answer

For each value of theta, there are four solutions, two of which are the trivial zero solution. All solutions can be saved in a cell array and then you can figure out which ones to keep. Or you can choose the solutions you want inside the loop.
syms r
syms t
eq = r^2 == r^2*cos(t)^2 + ((283*r^2*cos(t)^2)/50 + (3481*r*cos(t))/2500)^2;
thetas = sym(pi)*[0: 1/20: 1/2];
for ii = 1:numel(thetas)
R{ii} = solve(subs(eq,t,thetas(ii)),r);
end
% for example
R{2}
ans = 

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