Combine sine and cosine to prove bounds

$r\sin(x-\alpha)=r\cos\alpha\sin x-r\sin\alpha\cos x$. Hence $r\cos\alpha=3$, $r\sin\alpha=4$, so $r=5$ and $\alpha=\tan^{-1}(4/3)\approx53.1^\circ$. Therefore $3\sin x-4\cos x=5\sin(x-\alpha)$. Keep the exact phase for the identity; use $53.1^\circ$ only as an approximation, with $x$ in degrees.

$-1\le\sin(x-\alpha)\le1$, so $-5\le3\sin x-4\cos x\le5$. Adding $5$ gives $0\le3\sin x-4\cos x+5\le10$.