<s>[INST] 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?
 [/INST] 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
$$3\sin x +\cos x =2\iff \cos x\left(3\tan x+1\right)=2 \iff 3\tan x+1=\pm 2\sqrt{1+\tan^2x}$$
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