# 2020 AMC 12A Problems/Problem 9

## Problem

How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ 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 $\cos(\tfrac{x}{2})$ $\tan(2x)$ has a period of $\tfrac{\pi}{2},$ asymptotes at $x = \tfrac{\pi}{4}+\tfrac{k\pi}{2},$ and zeroes at $\tfrac{k\pi}{2}$. It is positive from $(0,\tfrac{\pi}{4}) \cup (\tfrac{\pi}{2},\tfrac{3\pi}{4}) \cup (\pi,\tfrac{5\pi}{4}) \cup (\tfrac{3\pi}{2},\tfrac{7\pi}{4})$ and negative elsewhere.

cos $(\tfrac{x}{2})$ 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

Or you could see the points at which both graphs are positive or both are negative, again yielding 5 such areas. -hi13

edited by - annabelle0913

## Solution 2

To find the asymptotes of $\tan(2x)$ we consider the behaviour of $\tan(x)$ on $[0,4\pi]$. Then we see that there are five separate continuous parts of the graph splitting the plane into regions. Since $\cos(\frac{x}{2})$ is continuous it must intersect each of the five pieces of $\tan$ at least once. But since $\tan(2x)$ is increasing and $\cos(\frac{x}{2})$ is decreasing on the interval and continuous increasing functions and decreasing functions can intersect at most once, there are $\boxed{\textbf{E) }5}$ intersections. -codecow

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