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!
 A: 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 $$\sin(x + \arctan(-1/3)) = 2/\sqrt{10}.$$
A: Hint to another possible way:
$$
3 \sin x -2= \cos x \Rightarrow 3 \sin x -2=\sqrt{1-\sin^2 x}
$$
squaring you have a second degree equation in $\sin x$. Also in this case be careful to extraneous solutions that can be introduced by squaring. 
A: Hint:
$$3\sin x - \cos x = 2 \implies 3\tan x - 1 = 2\sec x$$
$$9\tan^2x - 6\tan x + 1 = 4\sec^2x = 4(1 + \tan^2x)$$
$$5\tan^2x - 6\tan x - 3 = 0$$
Just need to be careful about any extraneous solutions
A: If
$a \sin x + b \cos x = c
$,
divide by
$\sqrt{a^2+b^2}$
to get
$A \sin x + B \cos x = C$
where
$A^2+B^2 = 1$.
Choose $y$
so that
$A = \cos y$ and $B = \sin y$
(i.e., 
$\tan y = B/A$).
Then
$C
= \sin x \cos y + \cos x \sin y
= \sin(x+y)
$,
so
$x+y = \sin^{-1}(C)$
or
$x = \sin^{-1}(C)-y$
.
As is often the case,
nothing original here.
A: 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.
