In the given equation $a \cos{x}^2+b \cos{x}+c=0$, we substitute \[ \cos{x}=\pm \sqrt{\dfrac{1+\cos{2x}}{2} }\] obtaining \[ a \left(\dfrac{1+\cos{2x}}{2}\right) \pm b \sqrt{\dfrac{1+\cos{2x}}{2}+c=0 }\] which simplifies to \[ (1) a^2 \cos^2{2x}+(2a^2+4ac-2b^2)\cos{2x}+a^2+4ac+4c^2-2b^2=0 \] For $a=4, b=2, c=-1$, the given equation is \[ 4\cos^2{x}+2\cos{x}-1=0 \] and the left member of $(1)$ is $4(4\cos^2{2x}+2\cos{2x}-1)$, so that the quadratic equation for $\cos{2x}$ has coefficients proportional to those of the given equation.

By $ \cos{x^2}$, do they mean $ \cos^2{x}$?

hello, multiplying the equation by $ 4(a\cos(x)^2-b\cos(x)+c)$ we obtain $ 4a^2\cos(x)^4+2(4ac-2b^2)\cos(x)^2+4c^2=0$. Plugging in this term $ 2\cos(x)^2=1+\cos(2x)$ we get after some algebraic manipulation $ a^2\cos(2x)^2+(2a^2+4ac-2b^2)\cos(2x)+a^2+4ac-2b^2+4c^2=0$ For $ a=4,b=2$ and $ c=-1$ we have $ 4\cos(x)^2+2\cos(x)-1=0$ and $ 16\cos(2x)^2+8\cos(2x)-4=0$ which implies $ 4\cos(2x)^2+2\cos(2x)-1=0$. Sonnhard.

What does the question want us to do?

I don't get the simplification part, could someone do it step by step?

qntty wrote: By $ \cos{x^2}$, do they mean $ \cos^2{x}$? ($ \cos{x}$)$^2$=$ \cos^2{x}$ It's important to notice: $ \cos({x^2})$ does NOT equal $ \cos^2{x}$, but it is a common typing mistake, like in this case. The official IMO website has the problem transcribed as $\color{red}\cos^2 x$ as you suggest, so I've edited it. Thanks! ~dj

Clearly, $\cos x = \sqrt{ \frac{\cos(2x)+1}{2}}$. Therefore, $$a\left(\frac{\cos (2x) +1}{2} \right) + b\sqrt{\frac{\cos (2x) + 1}{2}} + c = 0 \implies b^2 \left(\frac{\cos (2x) + 1}{2}\right) = \left( c + a\left( \frac{\cos (2x) + 1}{2}\right) \right)^2 \implies 0 = \cos ^2 (2x) \cdot a + \cos(2x)\left( -2b^2+2a^2+4ac\right) + 4ac + a^2 + 4c^2 - 2b^2$$Plugging in $a=4,b=2, $ and $c=-1$, we have that $\cos x = \frac{-2 \pm \sqrt{4-4(4)(-1)}}{8} = \frac{-1 \pm \sqrt{5}}{4}$. Plugging this into our equation for $\cos (2x)$, we have that $\cos(2x) = \frac{-1 - \sqrt{5}}{4}$.

Let $\cos x=r,s$ be the roots of $a\cos^2x+b\cos x+c$. Since $\cos2x=2\cos^2x-1$ by the double-angle formula, we seek the quadratic with roots $2r^2-1$ and $2s^2-1$. Furthermore, because $r+s=-\tfrac{b}{a}$ and $rs=\tfrac{c}{a}$ by Vieta's Formulas, $r^2+s^2=(r+s)^2-2rs=\tfrac{b^2}{a^2}-\tfrac{2c}{a}=\tfrac{b^2-2ac}{a}$. Hence, $$(2r^2-1)+(2s^2-1)=2(r^2+s^2)-2=\frac{2b^2-4ac}{a^2}-2=\tfrac{2b^2-4ac-a^2}{2}$$and $$(2r^2-1)(2s^2-1)=4r^2s^2-2(r^2+s^2)+1=\frac{4c^2}{a^2}-\frac{2b^2-4ac}{a^2}+1=\frac{a^2-2b^2+4c^2+4ac}{a^2},$$so $2r^2-1$ and $2s^2-1$ are the roots of the quadratic $$t^2+\frac{2a^2-2b^2+4ac}{a^2}t+\frac{a^2-2b^2+4c^2-4ac}{a^2}=0,$$or, after replacing $t$ with $\cos2x$ and scaling by $a^2$, $$\boxed{a^2\cos^22x+(2a^2-2b^2+4ac)\cos2x+(a^2-2b^2+4c^2+4ac)=0}.$$When $(a,b,c)=(4,2,-1)$, we have $$a\cos^2x+b\cos x+c=4\cos^2x+2\cos x-1=0$$and $$4^2\cos^22x+(2\cdot4^2-2\cdot2^2+4^2)\cos2x+(4^2-2\cdot2^2+4\cdot1^2-4^2)=16\cos^22x+8\cos2x-4=0,$$which do indeed have the same roots in $\cos x$ and $\cos2x$, respectively, as each is a scalar multiple of the other. $\blacksquare$

To be honest, I am slightly confused about the problem: In general, the quadratic equation $a\cos(x)^2+b\cos(x)+c=0$ will have two solutions for $\cos(x)$ and hence two solutions $x \in [0,\pi]$. However, the corresponding quadratic equation in $\cos(2x)$ will be a degree-4 equation in $\cos(x)$ and hence will in general have four solutions for $\cos(x)$ and hence four solutions $x \in [0,\pi]$. So I fail to see how it could be even possible to produce such an equation for general $a,b,c$ "whose roots are the same as those of the original equation" and I doubt that the solutions presented in this tread as well as the official one work. (What they produce, in my opinion, is a equation whose solutions include those of the original equation.) Can anyone clear/confirm my doubt?