We know that
$-1\le \cos x\le1$ so we if we multiply 3 so we get $-3\le 3\cos x\le 3$ and we also know that $-1\le \sin x\le 1$ and and again we multiply 4 we get $-4\le 4\sin x\le 4$ and we can add these two inequality we get $-7\le 3\cos x - 4\sin x\le 7$ but how in the above inequality is forming i m not able to understand at all.
Note that\begin{align}3\cos x-4\sin x&=5\left(\frac35\cos x-\frac45\sin x\right)\\&=5\left(\cos\alpha\cos x-\sin\alpha\sin x\right),\end{align}where $\alpha\in\Bbb R$ is such that $\cos\alpha=\frac35$ and that $\sin\alpha=\frac45$. Therefore$$3\cos x-4\sin x=5\cos(\alpha+x)\in[-5,5].$$
Hint : \begin{eqnarray*} 3 \cos \theta -4 \sin \theta = R \cos ( \theta +\phi) . \end{eqnarray*} $R=?$
You made a mistake adding the two inequalities when $3 cos(x)- 4 sin(x)$ requires that you subtract the two.
