Difference between revisions of "1961 IMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
| − | {{ | + | Since <math>cos^2x + sin^2x = 1</math>, we cannot have solutions with <math>n\ne2</math> and <math>0<|cos(x)|,|sin(x)|<1</math>. Nor can we have solutions with <math>n=2</math>, because the sign is wrong. So the only solutions have <math>sin (x) = 0</math> or <math>cos (x) = 0</math>, and these are: <math>x =</math> multiple of <math>\pi</math>, and <math>n</math> even; <math>x </math> even multiple of <math>\pi</math> and <math>n</math> odd; <math>x</math> = even multiple of <math>\pi + \frac{3\pi}{2}</math> and <math>n</math> odd. |
{{IMO box|year=1961|num-b=2|num-a=4}} | {{IMO box|year=1961|num-b=2|num-a=4}} | ||
Latest revision as of 00:07, 29 December 2007
Problem
Solve the equation
where 
is a given positive integer.
Solution
Since 
, we cannot have solutions with 
and 
. Nor can we have solutions with 
, because the sign is wrong. So the only solutions have 
or 
, and these are: 
multiple of 
, and even; 
even multiple of and odd; = even multiple of 
and odd.
| 1961 IMO (Problems) • Resources | ||
| Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
| All IMO Problems and Solutions | ||
