Correct Answer: $\sqrt{21}$

Solution : Given: $3 \sin x+4 \cos x=2$ (equation 1)
Let $3 \cos x–4 \sin x=k$ (equation 2)
We know the trigonometric identity, $\sin^2 x+ \cos^2 x=1$.
Squaring the equation and adding them together,
$(3 \cos x–4 \sin x)^2+(3 \sin x+4 \cos x)^2=k^2+2^2$
⇒ $9 \cos^2 x+16 \sin^2 x–24\sin x \cos x+9 \sin^2 x+16 \cos^2 x+24 \sin x \cos x=k^2+4$
⇒ $9( \cos^2 x+ \sin^2 x)–24\sin x \cos x+16 (\sin^2 x+ \cos^2 x)+24 \sin x \cos x=k^2+4$
⇒ $9+16=k^2+4$
⇒ $25=k^2+4$
⇒ $25–4=k^2$
⇒ $k^2=21$
⇒ $k=\sqrt{21}$
The value of $3 \cos x–4 \sin x=\sqrt {21}$.
Hence, the correct answer is $\sqrt {21}$.