Write the interval in which the value of

Let $f(x)=5 \cos x+3 \cos \left(x+\frac{\pi}{3}\right)+3$

$=5 \cos x+3\left(\cos x \cos 60^{\circ}-\sin x \sin 60^{\circ}\right)+3$

$=5 \cos x+\frac{3}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+3$

$=\frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+3$

We know that,

$-\sqrt{\left(\frac{13}{2}\right)^{2}+\left(\frac{3 \sqrt{3}}{2}\right)^{2}} \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x \leq \sqrt{\left(\frac{13}{2}\right)^{2}+\left(\frac{3 \sqrt{3}}{2}\right)^{2}}$

$-\sqrt{\frac{169}{4}+\frac{27}{4}} \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x \leq \sqrt{\frac{169}{4}+\frac{27}{4}}$

$\Rightarrow-\frac{14}{2} \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x \leq \frac{14}{2}$

$\Rightarrow-7+3 \leq \frac{13}{2} \cos x-\frac{3 \sqrt{3}}{2} \sin x+3 \leq 7+3$

Hence, $f(x)$ lies in the interval $[-4,10]$.