IMO 1962 - Problem 4

Problem: Solve the equation: $cos^2(x)+cos^2(2x)+cos^2(3x)=1$.
Solution: Since $cos(2x)=1+cos^2(x)$ and $cos(\alpha)+cos(\beta)=2cos(\frac{\alpha+\beta}{2})cos(\frac{\alpha-\beta}{2}),$ we have $cos^2(x)+cos^2(2x)+cos^2(3x)=1\iff cos(2x)+cos(4x)+2cos^2(3x)=2cos(3x)(cos(x)+cos(3x))=0\iff 4cos(3x)cos(2x)cos(x)=0$. Hence the solutions are $x\in \left\{\frac{\pi}{2}+m\pi,\frac{\pi}{4}+\frac{m\pi}{2},\frac{\pi}{6}+\frac{m\pi}{3}|m\in \mathbb{Z}\right\}$.