Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2020 AMC 12A Problems/Problem 9: Difference between revisions

← Older editNewer edit →
Quacker88 (talk | contribs)
editor
151 edits
P groudon (talk | contribs)
editor
421 edits
No edit summary
Line 1: Line 1:
==Problem==
==Problem==


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


<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>

Revision as of 06:39, 2 February 2020

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$(\frac{x}{2})$

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)

$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$. Applying double angle identities for both, we have

\[\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin x \cos x}{2\cos^{2}x-1}\]

Applying half angle identities on the RHS, we have $\cos\frac{x}{2}=\pm\sqrt{\frac{\cos x +1}{2}}$.

Setting both sides equal and squaring,

\[\frac{2\sin x \cos x}{2\cos^{2}x-1}=\pm\sqrt{\frac{\cos x + 1}{2}}\]

\[\frac{4\sin^2 x \cos^2 x}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}\]

Since $\sin^2 x + \cos^2 x = 1$, we can substitute $\sin^2 x = 1-\cos^2 x$ to convert the whole equation into cosine.

\[\frac{4(1-\cos^2 x) (\cos^2 x)}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}\]

Cross multiplying, we get

\[8(1-\cos^2 x) (\cos^2 x)=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)\]

\[0=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)-8(1-\cos^2 x) (\cos^2 x)\]

Without expanding anything, we can see that the first two polynomials will expand into a polynomial with degree $5$ and the $8(1-\cos^2 x) (\cos^2 x)$ term will expand into a polynomial with degree $4$. This means that overall, the polynomial will have degree $5$. From this, we can see that there are $\boxed{\textbf{E) }5}$ solutions. ~quacker88

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_9&oldid=116596"