If $\frac {9\sin x-3\cos x}{2\sin x + \cos x}=2$, define $\tan x$ and angle $x$. If: $$\frac {9\sin x-3\cos x}{2\sin x + \cos x}=2$$ $(0^\circ < x < 90^\circ$), define $\tan x$ and angle $x$.
Because of this in the parenthesis I know that I have to use some relation between complementary angles ex. $3\cos x = 3\sin(90^\circ-x)$ but when I include that in my problem I don't see any progress? I also don't really know what to do with the $2$ on the RHS. Any hints?

\begin{align}
& \frac {9\sin x-3\cos x}{2\sin x + \cos x}=2 \\[10pt]
\text{Therefore } & \frac{9\tan x - 3}{2\tan x + 1} = 2 \\[10pt]
& \text{and so on.}
\end{align}
