Difference between revisions of "2021 AMC 12B Problems/Problem 13"
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We rearrange to get <cmath>5\cos3\theta = 3\sin\theta-1.</cmath> | We rearrange to get <cmath>5\cos3\theta = 3\sin\theta-1.</cmath> | ||
We can graph two functions in this case: <math>y=5\cos{3x}</math> and <math>y=3\sin{x} -1 </math>. | We can graph two functions in this case: <math>y=5\cos{3x}</math> and <math>y=3\sin{x} -1 </math>. | ||
− | Using transformation of functions, we know that <math>5\cos{3x}</math> is just a cosine function with amplitude <math>5</math> and period <math>\frac{2\pi}{3}</math>. Similarly, <math>3\sin{x} -1 </math> is just a sine function with amplitude <math>3</math> and shifted <math>1</math> unit | + | Using transformation of functions, we know that <math>5\cos{3x}</math> is just a cosine function with amplitude <math>5</math> and period <math>\frac{2\pi}{3}</math>. Similarly, <math>3\sin{x} -1 </math> is just a sine function with amplitude <math>3</math> and shifted <math>1</math> unit downward: |
<asy> | <asy> | ||
import graph; | import graph; | ||
Line 22: | Line 22: | ||
add(legend(),point(E),20E,UnFill); | add(legend(),point(E),20E,UnFill); | ||
</asy> | </asy> | ||
− | + | So, we have <math>\boxed{\textbf{(D) }6}</math> solutions. | |
~Jamess2022 (burntTacos) | ~Jamess2022 (burntTacos) |
Revision as of 03:21, 28 January 2022
Contents
[hide]Problem
How many values of in the interval satisfy
Solution
We rearrange to get We can graph two functions in this case: and . Using transformation of functions, we know that is just a cosine function with amplitude and period . Similarly, is just a sine function with amplitude and shifted unit downward: So, we have solutions.
~Jamess2022 (burntTacos)
Video Solution by OmegaLearn (Using Sine and Cosine Graph)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.