Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$ IMO 1962/4

My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.

$$ \begin{align} \cos^22x&=(\cos^2x-\sin^2x)^2\\ &=\cos^4x+\sin^4x-2\sin^2\cos^2x\\ &=\cos^4+(1-\cos^2x)^2-2(1-\cos^2)\cos^2x\\ &=\cos^4+1-2\cos^2x+\cos^4x-2\cos^2x+2\cos^4x\\ &=4\cos^4x-4\cos^2x+1 \end{align} $$

Without knowledge of other trigonometric identities, $\cos3x$ can be derived using only Ptolemy's identities. However for the sake of brevity, let $\cos 3x=4\cos^3x-3\cos x$: $$ \begin{align} \cos^23x&=(4\cos^3x-3\cos x)^2\\ &=16\cos^6x+4\cos^2x-24\cos^4x \end{align} $$

Therefore, the original equation can be written as: $$\cos^2x+4\cos^4x-4\cos^2x+1+16\cos^6x+4\cos^2x-24\cos^4x-1=0$$ $$-20\cos^4x+6\cos^2x+16\cos^6x=0$$ Letting $y=\cos x$, we now have a polynomial equation: $$-20y^4+6y^2+16y^6=0$$ $$y^2(-20y^2+6y+16y^4)=0\Rightarrow y^2=0 \Rightarrow x=\cos^{-1}0=\bbox[yellow,10px]{90^\circ}$$ From one of the factors above, we let $z=y^2$, and we have the quadratic equation: $$16z^2-20z+6=0\Rightarrow 8z^2-10z+3=0$$ $$(8z-6)(z-\frac12)=0\Rightarrow z=\frac34 \& \ z=\frac12$$ Since $z=y^2$ and $y=\cos x$ we have: $$\biggl( y\rightarrow\pm\frac{\sqrt{3}}{2}, y\rightarrow\pm\frac{\sqrt{2}}2 \biggr)\Rightarrow \biggl(x\rightarrow\cos^{-1}\pm\frac{\sqrt{3}}{2},x\rightarrow\cos^{-1}\pm\frac{\sqrt{2}}2\biggr)$$ And thus the complete set of solution is: $$\bbox[yellow, 5px]{90^\circ, 30^\circ, 150^\circ, 45^\circ, 135^\circ}$$

As I do not have the copy of the answers, I still hope you can verify the accuracy of my solution.

But more importantly...

Seeing the values of $x$, is there a more intuitive and simpler way of finding $x$ that does away with the lengthy computation?

  • $\begingroup$ This post doesn't have the squares, but it could give some insight as to how one may think about this problem. $\endgroup$ – Arthur Mar 12 '18 at 13:12
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    $\begingroup$ This link gives you the solutions. $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 12 '18 at 13:15
  • $\begingroup$ I expect there to be a much more elegant solution to this as it is an IMO problem. Now only someone has to find it... $\endgroup$ – vrugtehagel Mar 12 '18 at 13:15
  • $\begingroup$ @vrugtehagel I believe the link to the solution is already an elegant answer! $\endgroup$ – John Glenn Mar 12 '18 at 13:17
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    $\begingroup$ Yup, I think so too! It was shortly posted before my comment, so I hadn't seen it. I advise @JoseArnaldoBebitaDris to summarize that solution and post it as answer, to avoid this question lingering in the unanswered section of this website $\endgroup$ – vrugtehagel Mar 12 '18 at 13:21

This is a summary of the solution found in this hyperlink.

We can write the LHS as a cubic function of $\cos^2 x$. This means that there are at most three values of $x$ that satisfy the equation.

Hence, we look for three values of $x$ that satisfy the equation and produce three distinct $\cos^2 x$. Indeed, we find that $$\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}$$ all satisfy the equation, and produce three different values for $\cos^2 x$, namely $0, \frac{1}{2}, \frac{3}{4}$.

Lastly, we solve the resulting equations $$\cos^2 x = 0$$ $$\cos^2 x = \frac{1}{2}$$ $$\cos^2 x = \frac{3}{4}$$ separately. We conclude that our solutions are: $$x=\frac{(2k+1)\pi}{2}, \frac{(2k+1)\pi}{4}, \frac{(6k+1)\pi}{6}, \frac{(6k+5)\pi}{6}, \forall k \in \mathbb{Z}.$$


You can shorten the argument by noting at the outset that $$ \cos3x=4\cos^3x-3\cos x=(4\cos^2x-3)\cos x $$ so if we set $y=\cos^2x$ we get the equation $$ y+(2y-1)^2+y(4y-3)^2=1 $$ When we do the simplifications, we get $$ 2y(8y^2-10y+3)=0 $$ The roots of the quadratic factor are $3/4$ and $1/2$.

A different strategy is to note that $\cos x=(e^{ix}+e^{-ix})/2$, so the equation can be rewritten $$ e^{2ix}+2+e^{-2ix}+e^{4ix}+2+e^{-4ix}+e^{6ix}+2+e^{-6ix}=4 $$ Setting $z=e^{2ix}$ we get $$ 2+z+z^2+z^3+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}=0 $$ or as well $$ z^6+z^5+z^4+2z^3+z^2+z+1=0 $$ that can be rewritten (noting that $z\ne1$), $$ \frac{z^7-1}{z-1}+z^3=0 $$ or $z^7+z^4-z^3-1=0$ that can be factored as $$ (z^3+1)(z^4-1)=0 $$ Hence we get (discarding the spurious root $z=1$) $$ 2x=\begin{cases} \dfrac{\pi}{3}+2k\pi \\[6px] \pi+2k\pi \\[6px] \dfrac{5\pi}{3}+2k\pi \\[12px] \dfrac{\pi}{2}+2k\pi \\[6px] \pi+2k\pi \\[6px] \dfrac{3\pi}{2}+2k\pi \end{cases} \qquad\text{that is}\qquad x=\begin{cases} \dfrac{\pi}{6}+k\pi \\[6px] \dfrac{\pi}{2}+k\pi \\[6px] \dfrac{5\pi}{6}+k\pi \\[6px] \dfrac{\pi}{4}+k\pi \\[6px] \dfrac{3\pi}{4}+k\pi \end{cases} $$

  • $\begingroup$ Great! An elegant solution too! $\endgroup$ – John Glenn Mar 12 '18 at 14:42


Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,




Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html


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