<s>[INST] Solutions to equation involving trigonometry and geometric series $$\cos^2 x + \cos ^3 x +\dots = 1+ \cos x$$ 
I want to find values of $x$ between $0$ and $ 180$ degrees for which the above equation holds true.
Attempt at a solution: left side is a converging geometric progression, for which $a_1$ is $\cos^2 x$ and $q = \cos x$. Plugging into the known formula for such series yields $\cos^2 x = 1$, which yields $45 $ and $135$ degrees as solutions. 
Is this okay? 
 [/INST] As mentioned, the LHS is a geometric series, with first term $ \cos^2 x $ and common ratio $ \cos x $, so it evaluates to $ \dfrac {\cos x}{1 - \cos x} $ and our equation becomes $$ \dfrac {\cos^2 x}{1-\cos x} = 1 + \cos x \implies \cos^2 x = 1 - \cos^2 x \implies \cos^2 x = \dfrac {1}{2}. $$Finally, take the square root of both sides and we have $ \cos x = \pm \dfrac {\sqrt{2}}{2} $ and our answers are $ \boxed {\dfrac{\pi}{4}, \dfrac{3\pi}{4}} $. 
Seems I've been beaten by a couple of minutes by ncmathsadist and a few seconds by Oliver!
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