26. Given the equation $a \cos ^{2} x+b=\cos x+c=0$, where $a, b$ and $c$ are given real numbers. Form a quadratic equation in terms of $a, b$, and $c$ whose roots are the corresponding values of $\cos 2 x$. Compare the given and the formed equations for the case $a=4, b=2, c=-1$.
26. We solve the given equation with respect to $\cos x$. Then we calculate $\cos 2 x$, using the formula $\cos 2 x=2 \cos ^{2} x-1$,

$$
\cos 2 x=\frac{b^{2}-2 a c \pm b \sqrt{b^{2}-4 a c}-a^{2}}{a^{2}}
$$

Using these values as roots, we form a new equation

$$
a^{2} \cos ^{2} 2 x+2\left(a^{2}+2 a c-b^{2}\right) \cos 2 x+(a+2 c)^{2}-2 b^{2}=0 .
$$

We notice that when $a=4, b=2, c=-1$, the coefficients of the given and the obtained equations are the same.