The minimum value of 3 cos x + 4 sin x + 8 is

$3 \cos x+4 \sin x+8$

first express $3 \cos x+4 \sin x$ as a $\cos (x+A)$

$3 \cos x+4 \sin x=a[\cos x \cos A-\sin x \sin A]$

i. e $3 \cos x+4 \sin x=a \cos x \cos A-a \sin x \sin A$

Now, equate the coefficients of $\sin x$ and $\cos x$ we get,

$a \cos A=3$

and $-a \sin A=4$

i. e $\frac{a \sin A}{a \cos A}=\frac{-4}{3}$

i. e $\tan A=\frac{-4}{3}$

i. e $A=\tan ^{-1}\left(\frac{-4}{3}\right)$

also, $(a \cos A)^{2}+(a \sin A)^{2}=9+16=25$

i. e $a^{2}=25$

i. e $a=5$

$\Rightarrow 3 \cos x+4 \sin x=5 \cos \left(x-\tan ^{-1}\left(\frac{-4}{3}\right)\right)$

i. e $3 \cos x+4 \sin x+8=5 \cos \left(x-\tan ^{-1}\left(\frac{-4}{3}\right)\right)+8$

Since minimum value of $\cos \theta=-1$

$\Rightarrow(3 \cos x+4 \sin x+8) \min =5(-1)+8$

i. e Minimum vlaue of $3 \cos x+4 \sin x+8=3$

Hence, the correct answer is option D.