If in a triangle $ABC,\cos 3A+\cos 3B+\cos 3C=1,$ then one angle must be exactly equal to 

We have,

      $\cos 3A+\cos 3B+\cos 3C=1$

$\begin{array}{r}\Rightarrow 2\cos 3\left(\frac{A+B}{2}\right)\cos 3\left(\frac{A-B}{2}\right)-\cos 3(A+B)=1\\ [\because C=\pi -(A+B\left)\right]\end{array}$

$\begin{array}{r}\Rightarrow 2\cos 3\left(\frac{A+B}{2}\right)\cos 3\left(\frac{A-B}{2}\right)-2{\cos }^{2}3\left(\frac{A+B}{2}\right)=0\\ \Rightarrow 2\cos 3\left(\frac{A+B}{2}\right)\left\{\cos 3\left(\frac{A-B}{2}\right)-\cos 3\left(\frac{A+B}{2}\right)\right\}=0\end{array}$

$\Rightarrow \cos 3\left(\frac{A+B}{2}\right)=0\text{ or, }\cos 3\left(\frac{A-B}{2}\right)=\cos 3\left(\frac{A+B}{2}\right)$

Now,

       $\cos 3\left(\frac{A+B}{2}\right)=0\Rightarrow A+B=\frac{\pi }{3}\Rightarrow C=\frac{2\pi }{3}$

and,

      $\cos 3\left(\frac{A-B}{2}\right)=\cos 3\left(\frac{A+B}{2}\right)\Rightarrow B=0,$ which is not possible.