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2024 AMC 12A Problems/Problem 8

Revision as of 09:17, 9 November 2024 by Thepowerful456 (talk | contribs)

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 1

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

Solution 1.1 (less words)

\[\log(\sin(3\theta))+\log(\cos(2\theta))=0\] \[\log(\sin(3\theta)\cos(2\theta))=0\] \[\sin(3\theta)\cos(2\theta)=1\] \[\text{Since } -1\le \sin(x),\cos(x)\le 1 \Rightarrow \sin(3\theta)=\cos(2\theta)= \pm1\] BUT note that $\log(-1)$ is not real \[\Rightarrow \sin(3\theta)=\cos(2\theta)= 1\] \[3\theta=\frac{\pi}{2}+2\pi n; \space 2\theta=2\pi m \space (m,n \in \mathbb{Z})\] \[\theta=\frac{\pi}{6}+\frac{2\pi n}{3}; \space \theta=\pi m\] \[\Rightarrow \theta\text { has no solution}\] Giving us $\fbox{(A) 0}$.

~Minor edits by evanhliu2009

Solution 2

Let $f(\theta)=\log(\sin(3\theta))$ and let $g(\theta)=\log(\cos(2\theta))$.

Note that $-1\leq\sin(3\theta),\cos(2\theta)\leq1$. Because the logarithm of a nonpositive number is not real, the functions $f$ and $g$ only exist when $\sin(3\theta)$ and $\cos(2\theta)$ are positive, respectively. Furthermore, because the logarithm of any positive real number less than $1$ is negative, the only case where the function $f+g$ could equal $0$ is if $f(\theta)=g(\theta)=0$, which only occurs when their respective sine and cosine expressions are both equal to $1$.

Thus, we have these two equations, where $m$ and $n$ are any integers: \begin{align*} \sin(3\theta)=1 \Rightarrow 3\theta = \frac\pi2+2\pi m &\Rightarrow \theta = \frac\pi6+\frac{2\pi}3 m = \frac{4m+1}6\pi\\ \cos(2\theta)=1 \Rightarrow 2\theta = 2\pi n &\Rightarrow \theta = \pi n \end{align*}

Because $m$ is an integer, $4m+1$ is odd, and so $\frac{4m+1}6$ is never an integer. Therefore, by the first equation, $\theta$ can never be an integer multiple of pi. Thus, because the second equation requires that $\theta$ be an integer multiple of pi, these two equations cannot both be satisfied, and so there are no solutions to the given equation.

Thus, we choose answer choice $\boxed{\textbf{(A) }0}$.

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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