I think we can use the fact that : cos²x = (cos2x+1)/2 I will post my solution as soon as possible!

Some ugly trigonometric computation later : $\cos^2{x}+\cos^2{2x}+\cos^2{3x} -1 = 2cos(x)^2 \times (2cos(x)^2-1) \times (4cos(x)^2-3)$ Is-it really IMO ?

Yes, this was an IMO question, the problems were trivial in the good old days

$\cos ^2 x + (\cos ^2 x - \sin ^2 x)^2 + ( - 3 + 4\cos ^3 x)^2 = 1 \\ \cos ^2 x + \cos ^4 x - 2\sin ^2 x\cos ^2 x + \sin ^4 x + 9\cos ^2 x - 24\cos ^4 x \\ + 16\cos ^6 x = 1 \\ \\ 10\cos ^2 x - 23\cos ^4 x + 16\cos ^6 x + \sin ^2 x(\sin ^2 x - 2\cos ^2 x) = 1 \\ 10\cos ^2 x - 23\cos ^4 x + 16\cos ^6 x + (1 - \cos ^2 x)(1 - \cos ^2 x - 2\cos ^2 x) = 1 \\ 10\cos ^2 x - 23\cos ^4 x + 16\cos ^6 x + 1 - 4\cos ^2 x + 3\cos ^4 x = 1 \\ 16\cos ^6 x - 20\cos ^4 x + 6\cos ^2 x = 0 \\ 2\cos ^2 x(3 - 10\cos ^2 x + 8\cos ^4 x) = 0 \\$

Solving the equations : $x=\frac{\pi }{2} +- k\pi (k=0,1,2,...)$ $x=\frac{\pi }{4} +- \frac{k\pi }{2} (k=0,1,2,...)$ $x=\frac{\pi }{6}+-k\pi (k=0,1,2,....)$ $x=\frac{5\pi }{6}+-k\pi (k=0,1,2,....)$

tµtµ wrote: Some ugly trigonometric computation computation? In fact, I was thinking about : (cos2x)² = (cos4x + 1)/2 (cos3x)² = (cos6x + 1)/2 So, if we sum : cos2x + cos4x + cos6x = -1 <==> -2cos2x + 2cos²2x + 4cos^3(2x) = 0 We can factor by 2cos2x, and this gives the solutions Am I misreasoning?

I forgot $cosx$ $\cos ^2 x + (\cos ^2 x - \sin ^2 x)^2 + ( - 3 + 4\cos ^3 x)^2 = 1$ $\cos ^2 x + (\cos ^2 x - \sin ^2 x)^2 + ( - 3cosx + 4\cos ^3 x)^2 = 1$

Much nicer solution:

Better $0=1+cos2x+cos4x+cos6x=Re(\sum_{k=0}^{3}e^{i2kx})=\frac{sin7x+sinx}{2sinx}=\frac{sin4xcos3x}{sinx}=4cosxcos2xcos3x$.

By the double- and triple-angle formulas, $\cos2x=2\cos^2x-1$ and $\cos3x=4\cos^3-3\cos x$, so we have $\cos^2x+(2\cos^2x-1)^2+(4\cos^3x-3\cos x)^2=1$. Let $k=\cos x$. Then, $$k^2+(2k^2-1)^2+(4k^3-3k)^2=k^2+(4k^4-4k^2+1)+(16k^6-24k^4+9k^2)=16k^6-20k^4+6k^2+1=1$$implying $$16k^6-20k^4+6k^2=2k^2(8k^4-10k^2+3)=2k^2(2k^2-1)(4k^2-3)=0.$$The distinct solutions to this equation are $k=-\tfrac{\sqrt{3}}{2},-\tfrac{\sqrt{2}}{2},0,\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{3}}{2}$. Hence, over the interval $x\in[0,\pi)$, the desired solutions to the original equation (noting $\cos x=k$) are $x=\boxed{\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{2},\tfrac{3\pi}{4},\tfrac{5\pi}{6}}$. If we do not restrict ourselves to $[0,\pi)$, we can add any multiple of $\pi$ to these angles. $\blacksquare$