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

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==Solution==
 
==Solution==
  
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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 cos<math>(\frac{x}{2})</math>
  
−
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)</math> and negative elsewhere.
+
tan<math>(2x)</math> has a period of <math>\frac{\pi}{2},</math> asymptotes at <math>x = \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)</math> and negative elsewhere.
  
−
<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)</math> and negative elsewhere.
+
cos<math>(\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)</math> and negative elsewhere.
  
 
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 19:16, 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 $x = \frac{\pi}{4}+\frac{k\pi}{2},$ 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 (Problems • Answer Key • Resources)
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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