Let $a,b,c$ be real numbers. Consider the quadratic equation in $\cos{x}$ \[ a \cos^2{x}+b \cos{x}+c=0.  \] Using the numbers $a,b,c$ form a quadratic equation in $\cos{2x}$ whose roots are the same as those of the original equation. Compare the equation in $\cos{x}$ and $\cos{2x}$ for $a=4$, $b=2$, $c=-1$.
1. Let $\cos x = r, s$ be the roots of the quadratic equation \(a \cos^2 x + b \cos x + c = 0\). By Vieta's formulas, we have:
   \[
   r + s = -\frac{b}{a} \quad \text{and} \quad rs = \frac{c}{a}.
   \]

2. Using the double-angle formula, \(\cos 2x = 2 \cos^2 x - 1\), we seek the quadratic equation whose roots are \(2r^2 - 1\) and \(2s^2 - 1\).

3. First, we need to find \(r^2 + s^2\). Using the identity \((r + s)^2 = r^2 + s^2 + 2rs\), we get:
   \[
   r^2 + s^2 = (r + s)^2 - 2rs = \left(-\frac{b}{a}\right)^2 - 2 \cdot \frac{c}{a} = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2}.
   \]

4. Now, we calculate the sum of the new roots \(2r^2 - 1\) and \(2s^2 - 1\):
   \[
   (2r^2 - 1) + (2s^2 - 1) = 2(r^2 + s^2) - 2 = 2 \left(\frac{b^2 - 2ac}{a^2}\right) - 2 = \frac{2(b^2 - 2ac)}{a^2} - 2 = \frac{2b^2 - 4ac - a^2}{a^2}.
   \]

5. Next, we calculate the product of the new roots \(2r^2 - 1\) and \(2s^2 - 1\):
   \[
   (2r^2 - 1)(2s^2 - 1) = 4r^2s^2 - 2(r^2 + s^2) + 1 = 4 \left(\frac{c}{a}\right)^2 - 2 \left(\frac{b^2 - 2ac}{a^2}\right) + 1 = \frac{4c^2}{a^2} - \frac{2(b^2 - 2ac)}{a^2} + 1.
   \]
   Simplifying further:
   \[
   \frac{4c^2 - 2b^2 + 4ac + a^2}{a^2}.
   \]

6. Therefore, the quadratic equation in \(\cos 2x\) with roots \(2r^2 - 1\) and \(2s^2 - 1\) is:
   \[
   t^2 + \frac{2b^2 - 4ac - a^2}{a^2} t + \frac{4c^2 - 2b^2 + 4ac + a^2}{a^2} = 0.
   \]
   Replacing \(t\) with \(\cos 2x\) and scaling by \(a^2\), we get:
   \[
   a^2 \cos^2 2x + (2b^2 - 4ac - a^2) \cos 2x + (4c^2 - 2b^2 + 4ac + a^2) = 0.
   \]

7. For \(a = 4\), \(b = 2\), and \(c = -1\), the original equation is:
   \[
   4 \cos^2 x + 2 \cos x - 1 = 0.
   \]

8. Substituting \(a = 4\), \(b = 2\), and \(c = -1\) into the quadratic equation in \(\cos 2x\), we get:
   \[
   16 \cos^2 2x + (2 \cdot 4^2 - 2 \cdot 2^2 + 4 \cdot 4 \cdot (-1)) \cos 2x + (4^2 - 2 \cdot 2^2 + 4 \cdot (-1)^2 + 4 \cdot 4 \cdot (-1)) = 0.
   \]
   Simplifying the coefficients:
   \[
   16 \cos^2 2x + (32 - 8 - 16) \cos 2x + (16 - 8 + 4 - 16) = 0,
   \]
   which simplifies to:
   \[
   16 \cos^2 2x + 8 \cos 2x - 4 = 0.
   \]

9. This equation is indeed a scalar multiple of the original equation, confirming that the roots in \(\cos x\) and \(\cos 2x\) are consistent.

The final answer is \(\boxed{16 \cos^2 2x + 8 \cos 2x - 4 = 0}\).