Let $3\sin x +\cos x =2 $ then $\frac{3\sin x}{4\sin x+3\cos x}=?$ Let $3\sin x +\cos x =2 $ then $\dfrac{3\sin x}{4\sin x+3\cos x}=\,?$

My try :
$$\frac{\frac{3\sin x}{\cos x}}{\frac{4\sin x+3\cos x}{\cos x}}=\frac{3\tan x}{4\tan x+3} =\;?$$
Now we have to find $\tan x$ from $3\sin x +\cos x =2 $ but how?

Instead of working with tangents, I'd recommend getting rid of the cosine and solving for $\sin x$:
$3\sin x+\cos x=2$ implies $1-\sin^2x=\cos^2x=(2-3\sin x)^2=4-12\sin x+9\sin^2x$, or
$$10\sin^2x-12\sin x+3=0$$
which solves to $\sin x=(6\pm\sqrt{36-30})/10=(6\pm\sqrt6)/10$. Both are valid solutions, since $|\sin x|=|6\pm\sqrt6|/10$ and $|\cos x|=|2-3\sin x|=|2\mp3\sqrt6|/10$ are less than or equal to $1$ for both signs.  
We now have
$${3\sin x\over4\sin x+3\cos x}={3\sin x\over4\sin x+3(2-3\sin x)}={3\sin x\over6-5\sin x}={3(6\pm\sqrt6)\over60-5(6\pm\sqrt6)}={3(6\pm\sqrt6)\over5(6\mp\sqrt6)}={3(6\pm\sqrt6)^2\over5\cdot30}\\={21\pm6\sqrt6\over25}$$
