Solving $\cos^2 x+\cos^2 2x +\cos^2 3x=1$ I picked this up from the IMO 1962.

Solve for $x$: 
  $$\cos^2 x+\cos^2 2x +\cos^2 3x=1$$ 

At first I thought it was trivial to bring this in an IMO, but I realized approaching it directly brings a power of 6 which is not all too friendly.
Is there a sneaky way to solve this?
 A: Use double angle identify to write the equation as
$$\cos 2x+\cos 4x +\cos 6x+1=0$$
Let $t=\cos 2x$. Then, the equation is
$$2t^3+t^2-t=0\implies t(t+1)(2t-1)=0$$
Thus, $\cos 2x= 0,\> -1, \> \frac12$ and the solutions are 
$x = \frac\pi4+\frac{n\pi}2,\> \frac\pi2+ n\pi,\> \pm\frac\pi6+\frac{2n\pi}3$.
A: Let $t:=\cos^2x$.
We have 
$$t+(2t-1)^2+(4t-3)^2t=1$$
or
$$16t^3-20t^2+6t=0.$$
The roots are $0,\dfrac12,\dfrac34$, nothing really difficult.
A: Use double angles to get the equivalent equation 
$$\cos(0x) + \cos(2x) + \cos(4x) + \cos(6x) = 0.$$
Pair (first and last term) and (middle ones) to get the equivalent form
$$\cos(x)\cos(2x)\cos(3x) = 0.$$
A: Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\cos^2x+\cos^22x=\cos^22x-\sin^2x+1=\cos(2x+x)\cos(2x-x)+1$$
Finally by http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$$\cos3x+\cos x=2\cos2x\cos x$$
A: It's $$\cos^2x\left(1+(4\cos^2x-3)^2\right)=4\sin^2x\cos^2x,$$ which gives $$\cos{x}=0$$ and $$x=90^{\circ}+180^{\circ}k,$$ where $k$ is an integer number, or $$1+(4\cos^2x-3)^2=4-4\cos^2x,$$ which is
$$8\cos^4x-10\cos^2x+3=0$$ or
$$(4\cos^2x-3)(2\cos^2x-1)=0$$ or
$$(2\cos2x-1)\cos2x=0,$$ which gives
$$x=\pm30^{\circ}+180^{\circ}k$$ or $$x=45^{\circ}+90^{\circ}k.$$
A: Hint:
use the identity 
$${\displaystyle \cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma =-2\cos \alpha \cos \beta \cos \gamma +1\,}$$
