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2020 AMC 12A Problems/Problem 9

Revision as of 16:13, 17 September 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution)

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

We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$

  1. The graph of $y=\tan(2x)$ has a period of $\frac{\pi}{2},$ asymptotes at $x=\frac{\pi}{4}+\frac{k\pi}{2},$ and zeros at $x=\frac{k\pi}{2}$ for some integer $k.$

    On the interval $[0,2\pi],$ the graph has five branches: \[\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].\] Note that $\tan(2x)\in[0,\infty)$ for the first branch, $\tan(2x)\in(-\infty,\infty)$ for the three middle branches, and $\tan(2x)\in(-\infty,0]$ for the last branch. Moreover, all branches are strictly increasing.

  2. The graph of $y=\cos\left(\frac x2\right)$ has a period of $4\pi$ and zeros at $x=\pi+2k\pi$ for some integer $k.$

    On the interval $[0,2\pi],$ note that $\cos\left(\frac x2\right)\in[-1,1].$ Moreover, the graph is strictly decreasing.

The graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right)$ intersect once on each of the five branches of $y=\tan(2x),$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(800,200); real f(real x) { return tan(2*x); } real g(real x) { return cos(x/2); } draw(graph(f,0,atan(3)/2),red,"$y=\tan(2x)$"); draw(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),red); draw(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),red); draw(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),red); draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red); draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$"); real xMin = 0; real xMax = 9/4*pi; real yMin = -3; real yMax = 3; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[], B[]; A[0] = (2pi,0); A[1] = (0,2); A[2] = (0,0); A[3] = (0,-2); B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0]; B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0]; B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0]; B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0]; B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0]; label("$2\pi$",A[0],(0,-2.5)); label("$2$",A[1],(-2.5,0)); label("$0$",A[2],(-2.5,0)); label("$-2$",A[3],(-2.5,0)); for (int i = 0; i < 5; ++i) { dot(B[i],black+linewidth(5)); } add(legend(),point(E),60E,UnFill); [/asy] Therefore, the answer is $\boxed{\textbf{(E)}\ 5}.$

~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow

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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