For example, in $\triangle ABC$, the measures of the three interior angles satisfy
$$
\frac{\angle A}{\angle B}=\frac{\angle B}{\angle C}=\frac{1}{3} \text {. }
$$

Find the value of $T=\cos A+\cos B+\cos C$. ${ }^{[6]}$
(2011, National High School Mathematics League Shanxi Province Preliminary Contest)
Let $\angle A=\theta$. Then $\angle B=3 \theta, \angle C=9 \theta$.
$$
\begin{array}{l}
\text { By } \theta+3 \theta+9 \theta=\pi \Rightarrow \theta=\frac{\pi}{13} \text {. } \\
\text { Hence } T=\cos \theta+\cos 3 \theta+\cos 9 \theta \\
=\cos \theta+\cos 3 \theta-\cos 4 \theta \\
=2 \cos \theta \cdot \cos 2 \theta-2 \cos ^{2} 2 \theta+1 \\
>1 \text {, } \\
T^{2}=(\cos \theta+\cos 3 \theta+\cos 9 \theta)^{2} \\
=\cos ^{2} \theta+\cos ^{2} 3 \theta+\cos ^{2} 9 \theta+2 \cos \theta \cdot \cos 3 \theta+ \\
2 \cos 3 \theta \cdot \cos 9 \theta+2 \cos 9 \theta \cdot \cos \theta \\
=\frac{1+\cos 2 \theta}{2}+\frac{1+\cos 6 \theta}{2}+\frac{1+\cos 18 \theta}{2}+ \\
(\cos 2 \theta+\cos 4 \theta)+(\cos 6 \theta+\cos 12 \theta)+ \\
(\cos 8 \theta+\cos 10 \theta) \text {. } \\
\end{array}
$$
$$
\begin{array}{l}
\text { And } T=\cos \theta+\cos 3 \theta+\cos 9 \theta \\
=-\cos 12 \theta-\cos 10 \theta-\cos 4 \theta \text {, } \\
\end{array}
$$

Then $2 T^{2}-T=3+3 \sum_{k=1}^{6} \cos 2 k \theta$.
$$
\text { Let } P=\sum_{k=1}^{6} \cos 2 k \theta \text {. }
$$

Hence $P \times 2 \sin \theta$
$$
\begin{array}{l}
=\sum_{k=1}^{6}[\sin (2 k+1) \theta-\sin (2 k-1) \theta] \\
=-\sin \theta .
\end{array}
$$

Therefore, $P=-\frac{1}{2}$.
Thus, $2 T^{2}-T=3-\frac{3}{2}=\frac{3}{2}$, which means $4 T^{2}-2 T-3=0$.
Since $T>1$, solving this equation gives $T=\frac{1+\sqrt{13}}{4}$.