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?
 A: Let $t=\tan(\frac x2)$.
use the identities
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2},$$
and
$$\tan(x)=\frac{2t}{1-t^2}.$$
thus
$$3\sin(x)+\cos(x)=2\implies$$
$$6t+1-t^2=2(1+t^2) \implies$$
or
$$3t^2-6t+1=0$$
hence
$$\frac{3\tan(x)}{4\tan(x)+3}=$$
$$\frac{6t}{8t+3(1-t^2)}=$$
$$\frac{6t}{2t+4}=3-\frac{6}{t+2}$$
with $$t=1\pm \sqrt{\frac 23}$$
A: 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
$${3\sin x\over4\sin x+3\cos x}={3\sin x\over4\sin x+3(2-3\sin x)}={3\sin x\over6-5\sin x}={3(6\pm\sqrt6)\over60-5(6\pm\sqrt6)}={3(6\pm\sqrt6)\over5(6\mp\sqrt6)}={3(6\pm\sqrt6)^2\over5\cdot30}\\={21\pm6\sqrt6\over25}$$
A: HINT
By tangent half-angle substitution with $t=\tan \frac x 2$ we have
$$3\sin x +\cos x =2 \iff 3\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}=2 \iff 3t^2-6t+1=0$$
then
$$\dfrac{3\sin x}{4\sin x+3\cos x}=\frac{6t}{-3t^2+8t+3}=\frac{6t}{(-3t^2-1)+8t+4}=\frac{6t}{2t+4}=\frac{3t}{t+2}$$
As an alternative since $\cos x=\pm \frac1{\sqrt{1+\tan^2x}}$ we can use that
$$3\sin x +\cos x =2\iff \cos x\left(3\tan x+1\right)=2 \iff 3\tan x+1=\pm 2\sqrt{1+\tan^2x}$$
