Difference between revisions of "2020 AMC 12A Problems/Problem 9"
Sevenoptimus (talk | contribs) (Removed a wrong solution - it makes the unjustified assumption that all 5 roots of the degree-5 polynomial will be valid values of cos x (i.e. will be between -1 and 1).) |
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Draw a graph of tan<math>(2x)</math> and cos<math>(\frac{x}{2})</math> | Draw a graph of tan<math>(2x)</math> and cos<math>(\frac{x}{2})</math> | ||
| − | tan<math>(2x)</math> has a period of <math>\frac{\pi}{2},</math> asymptotes at <math>x = \frac{\pi}{4}+\frac{k\pi}{2},</math> and zeroes at <math>\frac{k\pi}{2}</math>. It is positive from <math>(0,\frac{\pi}{4}) \cup (\frac{\pi}{2},\frac{3\pi}{4}) \cup (\pi,\frac{5\pi}{4}) \cup (\frac{ | + | tan<math>(2x)</math> has a period of <math>\frac{\pi}{2},</math> asymptotes at <math>x = \frac{\pi}{4}+\frac{k\pi}{2},</math> and zeroes at <math>\frac{k\pi}{2}</math>. It is positive from <math>(0,\frac{\pi}{4}) \cup (\frac{\pi}{2},\frac{3\pi}{4}) \cup (\pi,\frac{5\pi}{4}) \cup (\frac{3\pi}{2},\frac{7\pi}{4})</math> and negative elsewhere. |
cos<math>(\frac{x}{2})</math> has a period of <math>4\pi</math> and zeroes at <math>\pi</math>. It is positive from <math>[0,\pi)</math> and negative elsewhere. | cos<math>(\frac{x}{2})</math> has a period of <math>4\pi</math> and zeroes at <math>\pi</math>. It is positive from <math>[0,\pi)</math> and negative elsewhere. | ||
Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~lopkiloinm | Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~lopkiloinm | ||
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| + | edited by - annabelle0913 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 11:13, 17 February 2020
Contents
Problem
How many solutions does the equation 
have on the interval ![$[0,2\pi]?$](http://latex.artofproblemsolving.com/e/c/c/ecccefb4c460857891d99273237d9deb9ecf41c1.png)
Solution
Draw a graph of tan
and cos
tan has a period of 
asymptotes at 
and zeroes at 
. It is positive from 
and negative elsewhere.
cos has a period of 
and zeroes at 
. It is positive from 
and negative elsewhere.
Drawing such a graph would get 
~lopkiloinm
edited by - annabelle0913
Video Solution
~IceMatrix
See Also
| 2020 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
