<s>[INST] 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?

 [/INST] 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
</s>