Hello Luqman
Assuming N and n are set beforehand, there are four equations & five unknowns, so one of the unknowns gets to be the independent variable. It helps to pick an easy one, k1 and not theta or lambda. If you solve the first equation for theta in terms of k1, then everything else follows.
If theta is a solution, then theta +- 2*pi*m is also a solution for any integer m, so you have to decide how to deal with theta. The acot function restricts theta between -pi and pi, so if you run the initial code, you get jumps when theta goes past those boundaries. But staying within the boundaries is perfectly acceptable. If you go with the 'unwrap' line uncommented, the jumps are removed and the range of theta is larger. Valid solutions either way.
If the unwrap function is in, then theta is smooth and monotonic and you can use it as the independent variable, as in the second plot. I did not plot theta2 since theta1 + theta2 = n, making one of the two of them uninteresting. It looks like you can use lambda1 or lambda2 as the independent in some cases but not all.
If you are swapping around independent and dependent variables it helps to be dense in points. I used about a million since they're inexpensive.
N = 5;
n = 1;
k1 = 0:1e-5:8;
k2 = 2*pi*n/N -k1;
theta = 2*acot((1/2)*(cot(k1/2) - cot(k2/2)));
lam1 = (1/(2*pi))*(N*k1-theta);
lam2 = (1/(2*pi))*(N*k2+theta);
figure(1)
plot(k1,theta,k1,lam1)
figure(2)
plot(theta,k1,theta,lam1)
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