Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$\cos 3A+\cos 3B+\cos 3C-1=0\Rightarrow \cos 3A+\cos 3B+\cos 3C+\cos 3\pi =0$ $\Rightarrow 2\cos \frac{3A+3B}{2}\cos \frac{3A-3B}{2}+2\cos \frac{3\pi +3C}{2}\cos \frac{3\pi -3C}{2}=0$ Since $\frac{3A+3B}{2}=\frac{3\pi -3C}{2}$, this gives $2\cos \frac{3\pi -3C}{2}\left\{\cos \frac{3A-3B}{2}+\cos \frac{3\pi +3C}{2}\right\}=0$ $\Rightarrow -4\sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}=0$

So $\frac{3A}{2}=\pi$ or $\frac{3B}{2}=\pi$ or $\frac{3C}{2}=\pi$, i.e. $A=\frac{2\pi }{3}$ or $B=\frac{2\pi }{3}$ or $C=\frac{2\pi }{3}$