If 3 sin x + 4 cos x = 5,

(a) 0

$3 \sin x+4 \cos x=5$

$\frac{3}{5} \sin x+\frac{4}{5} \cos x=1$

Let $\cos \alpha=\frac{3}{5}$ and $\sin \alpha=\frac{4}{5}$.

$\therefore \cos \alpha \sin x+\sin \alpha \cos x=1$

$\Rightarrow \alpha+x=\frac{\pi}{2}$

$\Rightarrow x=\frac{\pi}{2}-\alpha$    ....(1)

We have to find the value of $4 \sin x-3 \cos x$.

$4 \sin \left(\frac{\pi}{2}-\alpha\right)-3 \cos \left(\frac{\pi}{2}-\alpha\right) \quad \ldots\{$ From eq $(1)\}$

$=4 \cos \alpha-3 \sin \alpha$

$=4 \times \frac{3}{5}-3 \times \frac{4}{5} \quad\left(\because \cos \alpha=\frac{3}{5}\right.$ and $\left.\sin \alpha=\frac{4}{5}\right)$

$=0$