Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 1.1 (less words))
Line 16: Line 16:
 
~Technodoggo
 
~Technodoggo
 
==Solution 1.1 (less words)==
 
==Solution 1.1 (less words)==
−
<cmath>log(sin(3\theta))+log(cos(2\theta))=0</cmath>
+
<cmath>\log(\sin(3\theta))+\log(\cos(2\theta))=0</cmath>
−
<cmath>log(sin(3\theta)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>\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>
+
<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
+
BUT note that <math>\log(-1)</math> is not real
−
<cmath>\Rightarrow sin(3\theta)=cos(2\theta)= 1</cmath>
+
<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>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>\theta=\frac{\pi}{6}+\frac{2\pi n}{3}; \space \theta=\pi m</cmath>
 
<cmath>\Rightarrow \theta\text { has no solution}</cmath>
 
<cmath>\Rightarrow \theta\text { has no solution}</cmath>
−
Giving us <math> \fbox{(A) 0}</math>
+
Giving us <math> \fbox{(A) 0}</math>.
 +
 
 +
~Minor edits by evanhliu2009
  
 
==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}}

Revision as of 21:44, 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 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12A_Problems/Problem_8&oldid=232368"