First, I think there is much more to trigonometry overall, than just this sort of problem (which is on a little bit the trickier side). If you look at a book, there will be plenty of trig that is much more foundational. So I think just saying "trig courses are different in 'murica...oooh la la" is not the right overall assessment, given the probable majority of content that's the same.
My copy of Schaum's precalculus says there's no hard and fast rule on how to solve different trig equations, but advocates trying the following three approaches:
A. Factoring (if already possible) into equation equal to zero.
B. Converting into a single function (e.g. getting rid of some cossq to have all sinsq in an equation).
C. Squaring both sides. They have the specific example sinx + cosx = 1, which is almost exactly yours (but ends up easy to factor). You "throw" cosx functions "onto the other side" and then square both sides, then replace sinsq with 1- cossq. Then group terms and you have a quadratic eqn in cosx. Really, it's not that different than if you substituted the radical sq(1-sinsqx) immediately for cosx and then cleaned it up by squaring after.
P.s. I kind of get the impression method 1 ends up the same as what Schaum's talked about. Or those (unlinked) videos you mentioned. I do think the methods 2 and 3 you showed are interesting. But unless you're deep in a strong trig course, you're not going to be jamming that sort of stuff all the time. Yeah...good to have it at the time. But as memory fades, not something you use all the time in higher courses.
P.s.s. I solved the problem using the Schaum's method and the quadratic method. Ans: x =arccos((-sqrt(3)+/-i)/2). Whatever the heck that means. I'm assuming the complex argument results for same reason as Daly not finding any answer. But if you'd given a different constant and the solution was in the reals, the method would have gotten there fine. And I didn't find it at all laborious to use. Maybe seven lines (and I like to show all the steps).