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:

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

1962 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
All IMO Problems and Solutions

Page

Toolbox

Search

1962 IMO Problems/Problem 4

Problem

Solution

See Also