Solving $3\sin x - \cos x = 2$ for $x \in [0, 2\pi)$ Problem:
Solve $$3\sin x - \cos x = 2, \ \ \  x \in [0, 2\pi)$$
My attempt:
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!

When you arrive at
$$
9\sin^2x - 6\sin x\cos x + \cos^2x = 4
$$
just note that $4=4\sin^2x+4\cos^2x$ so you can rewrite your equation as
$$
5\sin^2x-6\sin x\cos x-3\cos^2x=0
$$
Since $\cos x=0$ is not a solution, divide by $\cos^2x$ and get
$$
5\tan^2x-6\tan x-3=0
$$
so
$$
\tan x=\frac{3\pm\sqrt{24}}{5}
$$
You need to exclude the extraneous solution, of course.
A different method that doesn't need $\tan(x/2)$ is to set $X=\cos x$, $Y=\sin x$ and transform the equation into
$$
\begin{cases}
3Y-X=2\\
X^2+Y^2=1
\end{cases}
$$
so $X=3Y-2$ and then
$$
(3Y-2)^2+Y^2=1
$$
so
$$
10Y^2-12Y+3=0
$$
which gives $Y=\dfrac{6\pm\sqrt{6}}{10}$ and $X=\dfrac{-2\pm\sqrt{6}}{10}$ (with the same choice of signs). Thus $\tan x=Y/X$ and so
$$
\tan x=\frac{3\pm2\sqrt{6}}{5}
$$
as it is easy to verify. Note that this method is guaranteed not to introduce extraneous solutions.
