# 1962 IMO Problems/Problem 4

## Problem

Solve the equation $\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1$.

## Solution

First, note that we can write the left hand side as a cubic function of $\cos^2 x$. So there are at most $3$ distinct values of $\cos^2 x$ that satisfy this equation. Therefore, if we find three values of $x$ that satisfy the equation and produce three different $\cos^2 x$, then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that $\frac{\pi}2$, $\frac{\pi}4$, and $\frac{\pi}6$ all satisfy the equation, and produce three different values of $\cos^2 x$, namely $0$, $\frac12$, and $\frac34$. So we solve $\cos^2 x = \text{each of these}$. Therefore, our solutions are:

$x = \frac{(2k+1)\pi}2,\, \frac{(2k+1)\pi}4,\, \frac{(6k+1)\pi}6,\, \frac{(6k+5)\pi}6 \quad \forall k\in Z$