Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$ 
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?

 A: 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:
$$x=\frac{(2k+1)\pi}{2}, \frac{(2k+1)\pi}{4}, \frac{(6k+1)\pi}{6}, \frac{(6k+5)\pi}{6}, \forall k \in \mathbb{Z}.$$
A: 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 $z=1$)
$$
2x=\begin{cases}
\dfrac{\pi}{3}+2k\pi \\[6px]
\pi+2k\pi \\[6px]
\dfrac{5\pi}{3}+2k\pi \\[12px]
\dfrac{\pi}{2}+2k\pi \\[6px]
\pi+2k\pi \\[6px]
\dfrac{3\pi}{2}+2k\pi
\end{cases}
\qquad\text{that is}\qquad
x=\begin{cases}
\dfrac{\pi}{6}+k\pi \\[6px]
\dfrac{\pi}{2}+k\pi \\[6px]
\dfrac{5\pi}{6}+k\pi \\[6px]
\dfrac{\pi}{4}+k\pi \\[6px]
\dfrac{3\pi}{4}+k\pi
\end{cases}
$$
A: Hint:
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$0=\cos^2x+\cos^22x+\cos^23x-1$$
$$=\cos(3x+x)\cos(3x-x)+\cos^22x$$
$$=\cos2x(\cos4x+\cos2x)$$
Use 
http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
