<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] 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 $$t=1\pm \sqrt{\frac 23}$$
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