Difference between revisions of "2021 AMC 12B Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
| − | First, move terms to get <math>1+5\cos 3x=3\sin x</math>. After graphing, we find that there are <math>\boxed{6}</math> solutions (two in each period of <math>5\cos 3x</math>). -dstanz5 | + | First, move terms to get <math>1+5\cos 3x=3\sin x</math>. After graphing, we find that there are <math>\boxed{\textbf{(D) }6}</math> solutions (two in each period of <math>5\cos 3x</math>). -dstanz5 |
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==Solution 2== | ==Solution 2== | ||
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xaxis("$x$",BottomTop,LeftTicks); | xaxis("$x$",BottomTop,LeftTicks); | ||
yaxis("$y$",LeftRight,RightTicks(trailingzero)); | yaxis("$y$",LeftRight,RightTicks(trailingzero)); | ||
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add(legend(),point(E),20E,UnFill); | add(legend(),point(E),20E,UnFill); | ||
</asy> | </asy> | ||
| − | We have <math>\boxed{(D) 6}</math> solutions. ~Jamess2022 (burntTacos) | + | We have <math>\boxed{\textbf{(D) }6}</math> solutions. ~Jamess2022 (burntTacos) |
== Video Solution by OmegaLearn (Using Sine and Cosine Graph) == | == Video Solution by OmegaLearn (Using Sine and Cosine Graph) == | ||
Revision as of 18:47, 24 June 2021
Contents
Problem
How many values of
in the interval
satisfy![]()
Solution 1
First, move terms to get
. After graphing, we find that there are
solutions (two in each period of
). -dstanz5
Solution 2
We can graph two functions in this case:
and
.
Using transformation of functions, we know that
is just a cos function with
amplitude 5 and period
. Similarly,
is just a sin function
with amplitude 3 and shifted 1 unit downwards. 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.