IMO 1959 - Problem 3

Problem: Let $x$ be an angle and let the real number $a,b,c,cos(x)$ satisfy the following equation: $acos^2(x)+bcos(x)+c=0$.
Write the analogous quadratic equation for $a,b,c,cos(2x)$. Compare the given and the obtained equality for $a=4,b=2,c=-1$.
Solution: Multiplying the equality by $4(acos^2(x)-bcos(x)+c)$, we obtain $4a^2cos^4x+2(4ac-2b^2)cos^2(x)+4c^2=0$. Plugging in $2cos^2(x)=1+cos(2x)$ we obtain (after quite a bit of manipulation):
$a^2cos^2(2x)+(2a^2+4ac-2b^2)cos(2x)+(a^2+4ac-2b^2+4c^2)=0$.
For $a=4,b=2$ and $c=-1$ we get $4cos^2(x)+2cos(x)-1=0$ and $16cos^(2x)+8cos(2x)-4=0\implies 4cos^2(2x)+2cos(2x)-1=0$