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.
 A: We can take for example $a=-100$ and $b=100$.  However, it might be interesting to find sharp bounds $a$ and $b$. That is a standard max/min problem. We will solve the problem without using the calculus. 
Note that 
$$\cos(x-\pi/6)=\cos x\cos(\pi/6)+\sin x\sin(\pi/6)=(\cos x)(\sqrt{3}/2)+(\sin x)(1/2).$$
Thus 
$$3\cos x+5\cos(x-\pi/6)=\frac{6+5\sqrt{3}}{2}\cos x +\frac{5}{2}\sin x.$$
To make the structure clearer (and make typing easier), let $p=\frac{6+5\sqrt{3}}{2}$ and $q=\frac{5}{2}$.
Note that 
$$(p\cos x+q\sin x)^2+(p\sin x-q\cos x)^2=p^2+q^2.$$
Thus $(p\cos x+q\sin x)^2$ can never be bigger than $p^2+q^2$, and is equal to $p^2+q^2$ precisely if $p\sin x -q\cos x=0$, that is, if $\tan x=q/p$. 
In the first quadrant, $\tan x=q/p$  at roughly $0.33$ radians.  At that value of $x$, the numbers $\cos x$ and $\sin x$  are positive, so our function is maximized.
There is also a solution of $\tan x=q/p$ in the third quadrant, where $\cos x$ and $\sin x$ are both negative, so our function is minimized. Thus
$$-\sqrt{p^2+q^2} \le 3\cos x +5\cos(x-\pi/6) \le \sqrt{p^2+q^2},$$
and these bounds are best possible.
Comment: One can obtain the same result more mechanically by using the calculus.  The derivative of $p\cos x+q\sin x$ is $-p\sin x+q\cos x$. This derivative is $0$ when $\tan x=q/p$. 
Another way of looking at things is to rewrite our expression as 
$$\sqrt{p^2+q^2} \left(\frac{p}{\sqrt{p^2+q^2}}\cos x+\frac{q}{\sqrt{p^2+q^2}}\sin  x\right).$$
Let $\theta$ be the angle whose sine is $p/\sqrt{p^2+q^2}$ and whose cosine is $q/\sqrt{p^2+q^2}$.
Then our expression is equal to
$$\sqrt{p^2+q^2} \sin(x+\theta),$$
and sharp bounds follow easily.
A: 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 $b=-a=C=\sqrt{34+15\sqrt{3}}$.
