Let x be an angle and let the real numbers a,b,c,\text{cos}x satisfy the following equation:
a \text{cos}^{2}x+b\text{cos}x+c=0.
Write the analogous equation for a,b,c,\text{cos}2x. Compare the given and the obtained equality for a=4,b=2,c=-1.


One way of achieving the goal is to rewrite the equation as


and, after squaring, to use the identity \text{cos}2x=2\text{cos}x^{2}-1. Practically the same result could be obtained by first multiplying the given equation by a \text{cos}^{2}x-b\text{cos}x+c=0 and the carrying out the same substitution.

The result is the same in both cases:


Amazingly, for a=4,b=2,c=-1 this converts to exactly same equation. Both \text{cos}x and \text{cos}2x satisfy


Indeed, the latter equation is satisfied by

\text{cos}\frac{2\pi}{5}=\frac{\sqrt{5}-1}{4} and