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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(2x)=\frac{\sin(2x)}{\cos(2x)}</math>
+
<math>\tan(2x)=\frac{\sin(2x)}{\cos(2x)}</math>. Applying double angle identities for both, we have
 +
 
 +
<cmath>\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin x \cos x}{2\cos^{2}x-1}</cmath>
 +
 
 +
Applying half angle identities on the RHS, we have <math>\cos\frac{x}{2}=\pm\sqrt{\frac{\cos x +1}{2}}</math>.
 +
 
 +
Setting both sides equal and squaring,
 +
 
 +
<cmath>\frac{2\sin x \cos x}{2\cos^{2}x-1}=\pm\sqrt{\frac{\cos x + 1}{2}}</cmath>
 +
 
 +
<cmath>\frac{4\sin^2 x \cos^2 x}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}</cmath>
 +
 
 +
Since <math>\sin^2 x + \cos^2 x = 1</math>, we can substitute <math>\sin^2 x = 1-\cos^2 x</math> to convert the whole equation into cosine.
 +
 
 +
<cmath>\frac{4(1-\cos^2 x) (\cos^2 x)}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}</cmath>
 +
 
 +
Cross multiplying, we get
 +
 
 +
<cmath>8(1-\cos^2 x) (\cos^2 x)=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)</cmath>
 +
 
 +
<cmath>0=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)-8(1-\cos^2 x) (\cos^2 x)</cmath>
 +
 
 +
Without expanding anything, we can see that the first two polynomials will expand into a polynomial with degree <math>5</math> and the <math>8(1-\cos^2 x) (\cos^2 x)</math> term will expand into a polynomial with degree <math>4</math>. This means that overall, the polynomial will have degree <math>5</math>. From this, we can see that there are <math>\textbf{E) }5</math> solutions.
  
 
==See Also==
 
==See Also==

Revision as of 23:42, 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)

$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}$. 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 $\textbf{E) }5$ solutions.

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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