*Problem*

*Solution*

The problem brings to mind a variant of the Pythagorean theorem

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:

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

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:

and because of the symmetry,

which combines into a single formula:

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

Using the trigonometric formulas

we successively get

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,

*Answer*

Heuristics brought us pretty close but not quite. Here's the graph of function f(x):