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

Revision as of 09:17, 9 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 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 $\boxed{\textbf{(A) }0}$ .

@@ Line 28: / Line 28: @@
 ~Minor edits by evanhliu2009
+== Solution 2 ==
+Let <math>f(\theta)=\log(\sin(3\theta))</math> and let <math>g(\theta)=\log(\cos(2\theta))</math>.
+Note that <math>-1\leq\sin(3\theta),\cos(2\theta)\leq1</math>. Because the logarithm of a nonpositive number is not real, the functions <math>f</math> and <math>g</math> only exist when <math>\sin(3\theta)</math> and <math>\cos(2\theta)</math> are positive, respectively. Furthermore, because the logarithm of any positive real number less than <math>1</math> is negative, the only case where the function <math>f+g</math> could equal <math>0</math> is if <math>f(\theta)=g(\theta)=0</math>, which only occurs when their respective sine and cosine expressions are both equal to <math>1</math>.
+Thus, we have these two equations, where <math>m</math> and <math>n</math> 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 <math>m</math> is an integer, <math>4m+1</math> is odd, and so <math>\frac{4m+1}6</math> is never an integer. Therefore, by the first equation, <math>\theta</math> can never be an integer multiple of pi. Thus, because the second equation requires that <math>\theta</math> 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 <math>\boxed{\textbf{(A) }0}</math>.
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}}
 {{MAA Notice}}

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