# 1962 IMO, Problem 4

Problem

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

Solution

The problem brings to mind a variant of the Pythagorean theorem

\text{cos}^{2}\alpha+\text{cos}^{2}\beta+\text{cos}^{2}\gamma=1,

provided \alpha+\beta+\gamma=\pi. This leads to 6x=\pi and x=\pi/6. Since it is not necessary for the three angles x,2x,3x to add up to \pi, we need to account for a possible period. Each of the three squared cosines has a period of \pi, which gives:

x = \pi/6 + k\pi,

and, since the function f(x)=\text{cos}^{2}x+\text{cos}^{2}2x+\text{cos}^{2}3x is symmetric in the interval \left[0,\pi\right], we have to add another solution

x = \pi - \pi/6 + k\pi,

where k is any integer. Is that all? No! We have missed an almost obvious solution x=\pi/4 and, of course, the whole bunch of them:

x = \pi/4 + k\pi

and because of the symmetry,

x = \pi - \pi/4 + k\pi,

which combines into a single formula:

x = \pi/4 + k\pi/2.

Is that all? Hard to say. The heuristics was nice, but it does not help to learn whether we got a complete solution. So, let's start anew. Here's the equation

\text{cos}^{2}x+\text{cos}^{2}2x+\text{cos}^{2}3x=1.

Using the trigonometric formulas

\text{cos}2\alpha=2\text{cos}^{2}\alpha - 1 and then \text{cos}\alpha+\text{cos}\beta=2\text{cos}\frac{\alpha+\beta}{2}\text{cos}\frac{\alpha-\beta}{2}

we successively get

\text{cos}2x+1+\text{cos}4x+1+2\text{cos}^{2}3x=2,
2\text{cos}x \text{cos}3x + 2\text{cos}^{2}3x=0,
2\text{cos}3x (\text{cos}x + \text{cos}3x)=0,
\text{cos}3x \text{cos}2x \text{cos}x=0.

Thus we have three sets of solutions: \pi/2+k\pi, \pi/4+k\pi/2, \pi/6+k\pi/3 where we may observe that the third set overlaps the first one. So, finally,