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Difference between revisions of "2024 AMC 12A Problems/Problem 8"

(Created page with "How many angles <math>\theta</math> with <math>0\le\theta\le2\pi</math> satisfy <math>\log(\sin(3\theta))+\log(\cos(2\theta))=0</math>? <math> \textbf{(A) }0 \qquad \textbf{(...")
 
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==Problem==
 
How many angles <math>\theta</math> with <math>0\le\theta\le2\pi</math> satisfy <math>\log(\sin(3\theta))+\log(\cos(2\theta))=0</math>?
 
How many angles <math>\theta</math> with <math>0\le\theta\le2\pi</math> satisfy <math>\log(\sin(3\theta))+\log(\cos(2\theta))=0</math>?
  
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~Technodoggo
 
~Technodoggo
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==See also==
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{{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}}
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{{MAA Notice}}

Revision as of 17:47, 8 November 2024

Problem

How many angles $\theta$ with $0\le\theta\le2\pi$ satisfy $\log(\sin(3\theta))+\log(\cos(2\theta))=0$?

$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4 \qquad$

Solution

Note that this is equivalent to $\sin(3\theta)\cos(2\theta)=1$, which is clearly only possible when $\sin(3\theta)=\cos(2\theta)=\pm1$. (If either one is between $1$ and $-1$, the other one must be greater than $1$ or less than $-1$ to offset the product, which is impossible for sine and cosine.) They cannot be both $-1$ since we cannot take logarithms of negative numbers, so they are both $+1$. Then $3\theta$ is $\dfrac\pi2$ more than a multiple of $2\pi$ and $2\theta$ is a multiple of $2\pi$, so $\theta$ is $\dfrac\pi6$ more than a multiple of $\dfrac23\pi$ and also a multiple of $\pi$. However, a multiple of $\dfrac23\pi$ will always have a denominator of $1$ or $3$, and never $6$; it can thus never add with $\dfrac\pi6$ to form an integral multiple of $\pi$. Thus, there are $\boxed{\textbf{(A) }0}$ solutions.

~Technodoggo

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

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