If $1+\cos \alpha +{\cos }^2\alpha +\cdots =2-\sqrt{2},\text{ then }\alpha \text{ is equal to }(0<\alpha <\pi )$ 

Given, $1+\cos \alpha +{\cos }^2\alpha +\cdots =2-\sqrt{2}$   where $0<\alpha <\pi$

$\begin{array}{l}\Rightarrow \frac{1}{1-\cos \alpha }=2-\sqrt{2}[\because -1\le \cos \alpha \le 1]\\ \Rightarrow (2-\sqrt{2})(1-\cos \alpha )=1\\ \Rightarrow 2(1-\cos \alpha )-\sqrt{2}(1-\cos \alpha )=1\\ \Rightarrow 2-2\cos \alpha -\sqrt{2}+\sqrt{2}\cos \alpha =1\\ \Rightarrow (\sqrt{2}-2)\cos \alpha +2-\sqrt{2}=1\\ \Rightarrow \cos \alpha =\frac{1-2+\sqrt{2}}{\sqrt{2}-2}\\ \Rightarrow \cos \alpha =\frac{\sqrt{2}-1}{\sqrt{2}-2}\\ \Rightarrow \cos \alpha =\frac{(\sqrt{2}-1)(\sqrt{2}+2)}{(\sqrt{2}-2)(\sqrt{2}+2)}\\ \Rightarrow \cos \alpha =\frac{2+2\sqrt{2}-\sqrt{2}-2}{2-4}\\ \Rightarrow \cos \alpha =-\frac{1}{2}\left(\sqrt{2}\right)=-\frac{1}{\sqrt{2}}[\because \alpha \in (0,\pi \left)\right]\end{array}$