Supposed that we have the equation; $a\cos^2(x)+b\cos(x)+c=0$ in which $a,b,c$ are real numbers. Obviously, the equation is written in terms of $\cos(x)$. My question is if we could write it again in terms of $\cos(2x)$ with the same coefficients $a,b,c$ as above such that we have the same solutions as the original equation has? Thanks.
This second order Equation
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1Your question is not completely clear. Are you asking the third question from the very first International Math Olympiad? http://www.artofproblemsolving.com/Wiki/index.php/1959_IMO_Problems/Problem_3 – 2012-05-09
2 Answers
Do you mean you want, for the same $x$, to have both $a \cos^2(x) + b \cos(x) + c = 0$ and $a \cos^2(2x) + b \cos(2x) + c = 0$? Then, since $\cos(2x) = 2\cos^2(x) - 1$, if we write $t = \cos(x)$ we have $a t^2 + b t + c = 0$ and $a (2 t^2 - 1)^2 + b (2 t^2 - 1) + c = 0$. Eliminating $c$ by subtracting one from the other and simplifying, this becomes $\left( 2\,t+1 \right) \left( t-1 \right) \left( 2\,a{t}^{2}+at-a+b \right) = 0$.
Thus either $\cos(x) = -1/2$ or $\cos(x) = 1$ (those being the two cases where $\cos(2x) = \cos(x)$) or $b = a(2 t^2 + t - 1)$. In the latter case, $c = -a t^2 - b t$ and the last equation becomes $2\, \left( t-1 \right) \left( 2\,t+1 \right) \left( 2\,t-1 \right) \left( t+1 \right) a = 0$, which says either $a=0$ (and then $b=0$ and $c=0$), or $t=1$ or $t=-1/2$ (and then as before, $\cos(2x) = \cos(x)$), or $t=-1$ with $b = 0$ and $c = -a$ (i.e. if $\cos(x) = -1$, $\cos(2x) = 1$ and these both satisfy $t^2 - 1 = 0$), or $t=1/2$ with $b=0$ and $c=-a/4$ (i.e. if $\cos(x) = 1/2$, $\cos(2x) = -1/2$ and these both satisfy $t^2 - 1/4 = 0$).
You can't write it in terms of $\cos (2x)$ with the same $a, b, c$ as the original as it would imply $\cos (2x)= \cos x$. You can define $u=\frac x2$, then you have $a\cos^2 (2u)+b \cos (2u) +c =0$ but that doesn't feel like what you are looking for. You can then expand it as J. M. say. Maybe this will help clarify the request.