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Difference between revisions of "1962 IMO Problems/Problem 4"

(New page: ==Problem== Solve the equation <math>cos^2{x}+cos^2{2x}+cos^2{3x}=1</math>. ==Solution== {{solution}} ==See Also== {{IMO box|year=1962|num-b=3|num-a=5}})
 
(→‎Problem)
 
(2 intermediate revisions by 2 users not shown)
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==Problem==
 
==Problem==
−
Solve the equation <math>cos^2{x}+cos^2{2x}+cos^2{3x}=1</math>.
+
Solve the equation <math>\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1</math>.
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/bRza0zea-e0?si=_E-hIXLt05qvNY9r
 +
[Video Solution by little-fermat]
  
 
==Solution==
 
==Solution==
−
{{solution}}
+
First, note that we can write the left hand side as a cubic function of <math>\cos^2 x</math>. So there are at most <math>3</math> distinct values of <math>\cos^2 x</math> that satisfy this equation. Therefore, if we find three values of <math>x</math> that satisfy the equation and produce three different <math>\cos^2 x</math>, then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that <math>\frac{\pi}2</math>, <math>\frac{\pi}4</math>, and <math>\frac{\pi}6</math> all satisfy the equation, and produce three different values of <math>\cos^2 x</math>, namely <math>0</math>, <math>\frac12</math>, and <math>\frac34</math>. So we solve <math>\cos^2 x = \text{each of these}</math>. Therefore, our solutions are:
 +
 
 +
<math>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</math>
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1962|num-b=3|num-a=5}}
 
{{IMO box|year=1962|num-b=3|num-a=5}}

Latest revision as of 00:37, 4 September 2023

Problem

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

Video Solution

https://youtu.be/bRza0zea-e0?si=_E-hIXLt05qvNY9r [Video Solution by little-fermat]

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:

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

See Also

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