My proof is given below :
$\displaystyle a\cos^{2} x\ +\ b\cos x\ +\ c\ =\ 0$ is the given equation. We know, $\displaystyle \cos 2x\ =\ 2\cos^{2\ } x\ -\ 1$ and from this we can express $\displaystyle \cos x$ as $\displaystyle \cos x\ =\ \sqrt{\frac{1\ +\ \cos 2x}{2}} \ $.
\begin{gather*} a\ \frac{1\ \ +\ \cos 2x}{2} \ +\ b\sqrt{\frac{1\ +\ \cos 2x}{2}} \ +\ c\ =\ 0\\ \Longrightarrow \ b^{2} \ .\ \frac{1\ \ +\ \cos 2x}{2} \ =\ a^{2} \ \left(\frac{1\ \ +\ \cos 2x}{2}\right)^{2} +\ c^{2} \ +\ 2ac\ .\ \frac{1\ \ +\ \cos 2x}{2}\\ \Longrightarrow \ 2b^{2} \ .\ ( 1\ \ +\ \cos 2x) \ =\ a^{2} \ ( 1\ \ +\ \cos 2x)^{2} +\ 4c^{2} \ +\ 4ac\ .\ ( 1\ \ +\ \cos 2x)\\ \Longrightarrow \ a^{2} \ \cos^{2} \ 2x\ +\ \left( a^{2} \ +\ 4c^{2} \ +\ 4ac\ -\ 2b^{2}\right) \ +\ \left( 4ac\ +\ 2a^{2} \ -\ 2b^{2}\right)(\cos 2x) \ =\ 0\\ \end{gather*}