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

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</math>
 
</math>
  
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==Solution==
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==Solution 1==
  
 
Note that this is equivalent to <math>\sin(3\theta)\cos(2\theta)=1</math>, which is clearly only possible when <math>\sin(3\theta)=\cos(2\theta)=\pm1</math>. (If either one is between <math>1</math> and <math>-1</math>, the other one must be greater than <math>1</math> or less than <math>-1</math> to offset the product, which is impossible for sine and cosine.) They cannot be both <math>-1</math> since we cannot take logarithms of negative numbers, so they are both <math>+1</math>. Then <math>3\theta</math> is <math>\dfrac\pi2</math> more than a multiple of <math>2\pi</math> and <math>2\theta</math> is a multiple of <math>2\pi</math>, so <math>\theta</math> is <math>\dfrac\pi6</math> more than a multiple of <math>\dfrac23\pi</math> and also a multiple of <math>\pi</math>. However, a multiple of <math>\dfrac23\pi</math> will always have a denominator of <math>1</math> or <math>3</math>, and never <math>6</math>; it can thus never add with <math>\dfrac\pi6</math> to form an integral multiple of <math>\pi</math>. Thus, there are <math>\boxed{\textbf{(A) }0}</math> solutions.  
 
Note that this is equivalent to <math>\sin(3\theta)\cos(2\theta)=1</math>, which is clearly only possible when <math>\sin(3\theta)=\cos(2\theta)=\pm1</math>. (If either one is between <math>1</math> and <math>-1</math>, the other one must be greater than <math>1</math> or less than <math>-1</math> to offset the product, which is impossible for sine and cosine.) They cannot be both <math>-1</math> since we cannot take logarithms of negative numbers, so they are both <math>+1</math>. Then <math>3\theta</math> is <math>\dfrac\pi2</math> more than a multiple of <math>2\pi</math> and <math>2\theta</math> is a multiple of <math>2\pi</math>, so <math>\theta</math> is <math>\dfrac\pi6</math> more than a multiple of <math>\dfrac23\pi</math> and also a multiple of <math>\pi</math>. However, a multiple of <math>\dfrac23\pi</math> will always have a denominator of <math>1</math> or <math>3</math>, and never <math>6</math>; it can thus never add with <math>\dfrac\pi6</math> to form an integral multiple of <math>\pi</math>. Thus, there are <math>\boxed{\textbf{(A) }0}</math> solutions.  
  
 
~Technodoggo
 
~Technodoggo
 +
==Solution 1.1 (less words)==
 +
<cmath>\log(\sin(3\theta))+\log(\cos(2\theta))=0</cmath>
 +
<cmath>\log(\sin(3\theta)\cos(2\theta))=0</cmath>
 +
<cmath>\sin(3\theta)\cos(2\theta)=1</cmath>
 +
<cmath>\text{Since } -1\le \sin(x),\cos(x)\le 1 \Rightarrow \sin(3\theta)=\cos(2\theta)= \pm1</cmath>
 +
BUT note that <math>\log(-1)</math> is not real
 +
<cmath>\Rightarrow \sin(3\theta)=\cos(2\theta)= 1</cmath>
 +
<cmath>3\theta=\frac{\pi}{2}+2\pi n; \space 2\theta=2\pi m \space  (m,n \in \mathbb{Z})</cmath>
 +
<cmath>\theta=\frac{\pi}{6}+\frac{2\pi n}{3}; \space \theta=\pi m</cmath>
 +
<cmath>\Rightarrow \theta\text { has no solution}</cmath>
 +
Giving us <math> \fbox{(A) 0}</math>.
 +
 +
~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>.
 +
 +
== Solution 3 (solution 1.1 with graphing) ==
 +
<cmath>\log(\sin(3\theta))+\log(\cos(2\theta))=0</cmath>
 +
<cmath>\log(\sin(3\theta))=-\log(\cos(2\theta))=\log(\sec(2\theta))</cmath>
 +
<cmath>\therefore\text{let } y=\sin(3\theta)=\sec(2\theta), y > 0</cmath>
 +
Sketching the graphs <math>y=\sin(3x)</math> and <math>y=\sec(2x)</math> for <math>y>0</math>, we see there are no intersections.
 +
 +
Thus there are <math>\boxed{\textbf{(A) }0}</math> angles that work.
 +
 +
~oofitu
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2024|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:14, 10 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 $\boxed{\textbf{(A) }0}$.

Solution 3 (solution 1.1 with graphing)

\[\log(\sin(3\theta))+\log(\cos(2\theta))=0\] \[\log(\sin(3\theta))=-\log(\cos(2\theta))=\log(\sec(2\theta))\] \[\therefore\text{let } y=\sin(3\theta)=\sec(2\theta), y > 0\] Sketching the graphs $y=\sin(3x)$ and $y=\sec(2x)$ for $y>0$, we see there are no intersections.

Thus there are $\boxed{\textbf{(A) }0}$ angles that work.

~oofitu

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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