2006 AMC 12A Problems/Problem 15: Difference between revisions
m only one solution |
|||
| Line 6: | Line 6: | ||
== Solutions == | == Solutions == | ||
*For <math>\cos x = 0</math>, x must be in the form of <math>\frac{\pi}{2} + \pi n</math>, where <math>n</math> denotes any [[integer]]. | *For <math>\cos x = 0</math>, x must be in the form of <math>\frac{\pi}{2} + \pi n</math>, where <math>n</math> denotes any [[integer]]. | ||
*For <math>\cos (x+z) = 1 / 2</math>, <math>x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n</math>. | *For <math>\cos (x+z) = 1 / 2</math>, <math>x + z = \frac{\pi}{3} +2\pi n, \frac{5\pi}{3} + 2\pi n</math>. | ||
Revision as of 11:45, 4 August 2017
Problem
Suppose 
and 
. What is the smallest possible positive value of 
?
Solutions
- For
, x must be in the form of 
, where 
denotes any integer. - For
, 
.
, x must be in the form of 
, where 
denotes any 
, 
.The smallest possible value of will be that of 
.
See also
| 2006 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
