Solve $$3\sin x - \cos x = 2, \ \ \ x \in [0, 2\pi)$$
I am able to solve it using Weierstrass substitutions and a good bit of patience, but the problem was given at an exam at a level where such substitutions are not part of the curriculum.
I've tried to find ways to rewrite the equation by using $\sin x / \cos x = \tan x$ which I expect is the "desired" method, but for some reason, my algebra is failing me, and I can't seem to eliminate the $\sin$ and $\cos$ terms.
I've tried squaring both sides, but given the coefficient on $\sin x$, I can't find a way to use that identity either.
So far I end up with $$9\sin^2x - 6\sin x\cos x + \cos^2x = 4$$
Any help appreciated!