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

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==Solution (Algebraically)==
 
==Solution (Algebraically)==
  
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<math>\tan</math>
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<math>\tan(2x)=\frac{\sin(2x)}{\cos(2x)}</math>
  
 
==See Also==
 
==See Also==

Revision as of 22:30, 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 $(0,\frac{\pi}{4}) \cup (\frac{\pi}{2},\frac{3\pi}{4}) \cup (\pi,\frac{5\pi}{4}) \cup (\frac{7\pi}{4},2\pi)$ 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

Solution (Algebraically)

$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$

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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