IMO 1962 Problem 4

Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 21m14s

Problem

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

Exploration

Consider the equation

$\cos^2 x + \cos^2 2x + \cos^2 3x = 1.$

Using standard trigonometric identities, we have $\cos 2x = 2\cos^2 x - 1$ and $\cos 3x = 4\cos^3 x - 3\cos x$. Squaring these, we obtain $\cos^2 2x = 4\cos^4 x - 4\cos^2 x + 1$ and $\cos^2 3x = 16\cos^6 x - 24\cos^4 x + 9\cos^2 x$. Substituting these into the original equation gives a degree-six polynomial in $\cos x$:

$\cos^2 x + \cos^2 2x + \cos^2 3x = 16\cos^6 x - 20\cos^4 x + 6\cos^2 x + 1.$

Setting this equal to 1 leads to $16\cos^6 x - 20\cos^4 x + 6\cos^2 x = 0$, which factors as $\cos^2 x (16\cos^4 x - 20\cos^2 x + 6) = 0$. Substituting $y = \cos^2 x$, we solve $16y^2 - 20y + 6 = 0$ and obtain $y = 3/4$ and $y = 1/2$. Together with $\cos^2 x = 0$, the candidate values are $\cos^2 x \in {0, 1/2, 3/4}$.

Testing these small values against the original equation:

$\cos^2 x = 0$ gives $x = \pi/2 + k\pi$; substitution yields $\cos^2 2x + \cos^2 3x = 1$, consistent.
$\cos^2 x = 1/2$ gives $x = \pm \pi/4 + k\pi/2$; substitution yields $\cos^2 2x + \cos^2 3x = 1/2 + 0 = 1$, consistent.
$\cos^2 x = 3/4$ gives $x = \pm \pi/6 + 2k\pi$ or $\pm 5\pi/6 + 2k\pi$; substitution yields $\cos^2 2x + \cos^2 3x = 1/4 + 0 = 1$, consistent.

No extraneous roots arise because every step uses identities that are reversible.

Exploring these substitutions for small $k$ confirms the computations and ensures that the pattern holds for all integers $k$.

Problem Understanding

The problem asks to find all real numbers $x$ such that

$\cos^2 x + \cos^2 2x + \cos^2 3x = 1.$

This is a Type A problem, requiring both verification of every solution (existence) and proof that no other solutions exist (exhaustiveness). The equation involves multiple angles, suggesting that reducing it to a polynomial in $\cos x$ is a natural strategy. The critical step is ensuring that every admissible cosine value is converted into a complete family of real $x$ values and that no other possibilities are omitted.

Key Observations

Using double-angle and triple-angle identities reduces the equation to a polynomial in $\cos x$. Squaring and substitution do not introduce extraneous roots.
The polynomial factorization yields three distinct cases: $\cos^2 x = 0$, $\cos^2 x = 1/2$, and $\cos^2 x = 3/4$.
For each $\cos^2 x$, the corresponding set of real $x$ values is given by standard cosine solutions.
Verification for each case confirms that the sum of squares equals 1.
No other values of $\cos^2 x$ satisfy the polynomial, so all solutions are captured.

Solution

We begin by substituting $\cos 2x = 2\cos^2 x - 1$ and $\cos 3x = 4\cos^3 x - 3\cos x$ into the original equation. Squaring gives:

$\cos^2 2x = 4\cos^4 x - 4\cos^2 x + 1,$

$\cos^2 3x = 16\cos^6 x - 24\cos^4 x + 9\cos^2 x.$

Hence

$\cos^2 x + \cos^2 2x + \cos^2 3x = \cos^2 x + (4\cos^4 x - 4\cos^2 x + 1) + (16\cos^6 x - 24\cos^4 x + 9\cos^2 x) = 16\cos^6 x - 20\cos^4 x + 6\cos^2 x + 1.$

Setting this equal to 1 gives

$16\cos^6 x - 20\cos^4 x + 6\cos^2 x = 0,$

$\cos^2 x (16\cos^4 x - 20\cos^2 x + 6) = 0.$

Substituting $y = \cos^2 x$, the quadratic equation $16y^2 - 20y + 6 = 0$ has discriminant $\Delta = 400 - 384 = 16$, yielding roots $y = 3/4$ and $y = 1/2$. Together with $y = 0$, we have $\cos^2 x \in {0, 1/2, 3/4}$.

We now classify all corresponding $x$:

If $\cos^2 x = 0$, then $\cos x = 0$ and

$x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}.$

Substituting into the original equation:

$\cos^2 x + \cos^2 2x + \cos^2 3x = 0 + \cos^2 (\pi + 2k\pi) + \cos^2 (3\pi/2 + 3k\pi) = 0 + 1 + 0 = 1.$

Hence these are valid solutions.

If $\cos^2 x = 1/2$, then $\cos x = \pm 1/\sqrt{2}$ and

$x = \frac{\pi}{4} + k\frac{\pi}{2}, \quad k \in \mathbb{Z}.$

Check: $\cos 2x = 0$, $\cos 3x = -\cos x$ or $\cos x$, so $\cos^2 x + \cos^2 2x + \cos^2 3x = 1/2 + 0 + 1/2 = 1$. These are valid.

If $\cos^2 x = 3/4$, then $\cos x = \pm \sqrt{3}/2$ and

$x = \pm \frac{\pi}{6} + 2k\pi \quad\text{or}\quad x = \pm \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}.$

Check: $\cos 2x = 2\cdot 3/4 - 1 = 1/2$, so $\cos^2 2x = 1/4$, and $\cos 3x = 0$, so $\cos^2 3x = 0$. Also $\cos^2 x = 3/4$, sum: $3/4 + 1/4 + 0 = 1$. These are valid.

We have exhausted all roots of the polynomial, so no other solutions exist.

Therefore, the complete solution set is

$x = \frac{\pi}{2} + k\pi, \quad x = \frac{\pi}{4} + k\frac{\pi}{2}, \quad x = -\frac{\pi}{4} + k\frac{\pi}{2}, \quad x = \frac{\pi}{6} + 2k\pi, \quad x = -\frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad x = -\frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}.$

$$\boxed{ \left{,x \in \mathbb{R} : \cos x = 0, \cos x = \pm \frac{1}{\sqrt{2}}, \cos x = \pm \frac{\sqrt{3}}{2},\right}. }$$

Verification of Key Steps

The double-angle and triple-angle identities are standard and valid for all real $x$. Squaring preserves equality and produces a polynomial equation equivalent to the original. The factorization $\cos^