Prove that $$\color\red{-4}\leq5\cos\theta+3\cos(\theta+\frac{\pi}{3})+3\leq\color\red{10}$$
My attempt:-
I simplified the equation to
$$\begin{align} &\;\;\phantom{=} 5\cos\theta+3\cos(\theta+60^0)+3 \\[4pt] &= 5\cos\theta+3(\cos\theta\cos60^0-\sin\theta\sin60^0)+3\\[4pt] &= \frac{13}{2}\cos\theta-\frac{3\sqrt3}{2}\sin\theta+3 \end{align}$$
I find no way to continue and eliminate the $\sqrt3$ and get the exact values of $-4$ and $10$ although solving the above equation gives me a near approximation.
How do I get the exact value (best without calculus but it may be accepted)?
Thanks for any help!!