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Solution of trigonometric equation

If a triangle ABC,

Cos(3A)+Cos(3B)+Cos(3C)=1 ,

$\angle A + \angle B \lt \angle C $

Then find $\angle C $


Since $\angle A + \angle B \lt \angle C $

Since $ \angle A + \angle B +\angle C \lt 2\angle C $

$\therefore      2\angle C \gt π$

$\Rightarrow       \angle C \gt \cfrac{π}{2} $


$Cos(3A) + Cos(3B) = 1 - Cos(3C)$

$\cancel{2} Cos(\frac{3A+3B}{2})Cos(\frac{3A-3B}{2})=\cancel{2}Sin^2(\frac{3C}{2})$

$Cos(\frac{3π-(3C)}{2})Cos(\frac{3A-3B}{2}) $ $= Sin(\frac{3C}{2})Sin(\frac{3π-(3A+3B)}{2}) $

$ -Sin(\frac{3C}{2})(Cos(\frac{3A-3B}{2})-Cos(\frac{3A+3B}{2}))=0$

$2Sin(\frac{3C}{2})Sin(\frac{3A}{2})Sin(\frac{3B}{2})=0 $

$\Rightarrow$ at least one of angle $\cfrac{3A}{2}$, $\cfrac{3B}{2}$, and $\cfrac{3C}{2}$ is π.

And   $\cfrac{3A}{2} + \cfrac{3B}{2} + \cfrac{3C}{2} =\cfrac{3π}{2} $

Where $\angle  \cfrac{3C}{2} \gt \cfrac{3π}{4} $

Hence $ \angle  \cfrac{3C}{2} = π $

$\Rightarrow     \angle C = \cfrac{2π}{3} $

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