You CANNOT solve for both theta1 and theta2 in a single equation like that. (I assume you meant a single equation, because you had the word OR in between the pair of equations you showed.) At best, you can solve for one variable in terms of the other. But effectively, what I will do here will allow you to generate infinintely many solution pairs, and do so trivially.
By the way, learn to use an asterisk to denote multiplication. Here is one of the equations you posed as an example:
E = -4*sin(theta1) - 6*sin(theta2) + 5*sin(theta2) == -3;
We can now simply solve for theta1, if we knew theta2. That is, we can do:
So there are two solutions. At least they may be real solutions.
So the two solutions cross at a point. As long as you know theta2, you can compute TWO solutions for theta1. And then there would be infinitely more solutions at multiples of 2*pi up or down.
Essentially, you can pick ANY value for theta2, that gives you two primary solutions for the other. Any such equation will work the same way, although it may not be as trivial to solve. If you want to generate a list of solutions, then just pick a list for theta2...
theta2list = linspace(-pi,pi,100);
Now substitute those elements into the solution, computing a a pair of solutions for every theta2.
What I did above was little more complicated than if you had posed the problem
theta1 + 2*theta2 == 3
Surely you would know how to deal with that?