*Problem*

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.

*Solution*

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

b\text{cos}x=-a\text{cos}^{2}x-c

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:

a^{2}\text{cos}^{2}2x+(2a^{2}+4ac-2b^2)\text{cos}2x+(a^{2}+4ac-2b^{2}+4c^2)=0.

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

4y^{2}+2y-1=0.

Indeed, the latter equation is satisfied by

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

-\text{cos}\frac{2\pi}{10}=\frac{\sqrt{5}+1}{4}.