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!

Another approach is to write: $$A \sin(x) + B \cos(x)=C$$ and then divide by $$\sqrt{A^2 + B^2}$$ to get:
$$\frac{A}{\sqrt{A^2+B^2}}\sin(x) + \frac{B}{\sqrt{A^2+B^2}}\cos(x) = \frac{C}{\sqrt{A^2+B^2}}$$
Since $$\left(\frac{A}{\sqrt{A^2+B^2}}\right)^2 + \left(\frac{B}{\sqrt{A^2+B^2}}\right)^2 = 1$$ there is a $\psi$ such that $$\cos(\psi) = \frac{A}{\sqrt{A^2+B^2}} \text{ and } \sin(\psi) = \frac{B}{\sqrt{A^2+B^2}}.$$
Thus we have $$\cos(\psi)\sin(x) + \sin(\psi)\cos(x) = \frac{C}{\sqrt{A^2+B^2}}$$ and using the sum formula we find:
$$\sin(x+\psi) = \frac{C}{\sqrt{A^2+B^2}}.$$

In this case $A=3$ and $B=-1$ so $\sqrt{A^2+B^2} = \sqrt{10}$. Since $A > 0$ and $B < -1$, this means $\psi$ is in the third quadrant. Thus $\arctan(-1/3) = \psi$.
Finally this means $$\sin(x + \arctan(-1/3)) = 2/\sqrt{10}.$$
