2024 AMC 12A Problems/Problem 8

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

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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2024 AMC 12A Problems/Problem 8

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