<s>[INST] Problem based on Range Find $a$ and $b$ such that the inequality $a \le 3 \cos{x} + 5\cos\left(x - \frac{\pi}{6}\right) \le b$ holds good for all x.
 [/INST] Remark: In the meantime this approach has been also added  by André Nicolas to his answer. 
We will show that the sum $3\cos x+5\cos \left( x-\frac{\pi }{6}\right) $
can be written as $C\sin (x+\phi )$. From the difference formula for $\cos
\left( x-\frac{\pi }{6}\right) $ and using the values $\cos \frac{\pi }{6}=
\frac{1}{2}\sqrt{3}$ and $\sin \frac{\pi }{6}=\frac{1}{2}$ we get
$$
3\cos x+5\cos \left( x-\frac{\pi }{6}\right) =\left( 3+\frac{5}{2}\sqrt{3}
\right) \cos x+\frac{5}{2}\sin x.
$$
The general identity 
$$
\begin{eqnarray*}
A\cos x+B\sin x &=&C\sin (x+\phi ) \\
&=&C\sin \phi \cos x+C\cos \phi \sin x
\end{eqnarray*}
$$
holds if $C\sin \phi =A$, $C\cos \phi =B$ i.e. $C=\sqrt{A^{2}+B^{2}}$ and $
\phi =\arctan \frac{A}{B}$. For $A=3+\frac{5}{2}\sqrt{3}$ and $B=\frac{5}{2}$
it takes the form
$$
\left( 3+\frac{5}{2}\sqrt{3}\right) \cos x+\frac{5}{2}\sin x=\sqrt{34+15
\sqrt{3}}\sin \left( x+\arctan \left( \frac{6}{5}+\sqrt{3}\right) \right). 
$$
For $C>0$ the function $C\sin (x+\phi )\in \left[ -C,C\right] $.
Consequently the best narrow bounds are $b=-a=C=\sqrt{34+15\sqrt{3}}$.
</s>