Difference between revisions of "2020 AMC 12A Problems/Problem 9"

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Draw a graph of tan<math>(2x)</math> and <math>cos(\frac{x}{2})</math>
 
Draw a graph of tan<math>(2x)</math> and <math>cos(\frac{x}{2})</math>
  
tan<math>(2x)</math> has a period of <math>\frac{\pi}{2}</math> and asymptotes at <math>\frac{\pi}{4}+\frac{k\pi}{2}</math>.
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tan<math>(2x)</math> has a period of <math>\frac{\pi}{2},</math> asymptotes at <math>\frac{\pi}{4}+\frac{k\pi}{2},</math> and zeroes at <math>\frac{k\pi}{2}</math>. It is positive from <math>(0,\frac{\pi}{4}) \cup (\frac{\pi}{2},\frac{3\pi}{4}) \cup (\pi,\frac{5\pi}{4}) \cup (\frac{4\pi}{4},2\pi) and negative elsewhere.</math>
  
<math>cos(\frac{x}{2})</math> has a period of <math>4\pi</math> and zeroes at <math>\pi</math>.
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<math>cos(\frac{x}{2})</math> has a period of <math>4\pi</math> and zeroes at <math>\pi</math>. It is positive from <math>[0,\pi) and negative elsewhere.</math>
  
 
Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~lopkiloinm
 
Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~lopkiloinm

Revision as of 20:07, 1 February 2020

Problem

How many solutions does the equation tan$(2x)=cos(\frac{x}{2})$ have on the interval $[0,2\pi]?$

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution

Draw a graph of tan$(2x)$ and $cos(\frac{x}{2})$

tan$(2x)$ has a period of $\frac{\pi}{2},$ asymptotes at $\frac{\pi}{4}+\frac{k\pi}{2},$ and zeroes at $\frac{k\pi}{2}$. It is positive from $(0,\frac{\pi}{4}) \cup (\frac{\pi}{2},\frac{3\pi}{4}) \cup (\pi,\frac{5\pi}{4}) \cup (\frac{4\pi}{4},2\pi) and negative elsewhere.$

$cos(\frac{x}{2})$ has a period of $4\pi$ and zeroes at $\pi$. It is positive from $[0,\pi) and negative elsewhere.$

Drawing such a graph would get $\boxed{\textbf{E) }5}$ ~lopkiloinm

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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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