Difference between revisions of "2020 AMC 12A Problems/Problem 9"
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<cmath>0=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)-8(1-\cos^2 x) (\cos^2 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>\boxed{\textbf{E) }5}</math> solutions. | + | 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>\boxed{\textbf{E) }5}</math> solutions. ~quacker88 |
==See Also== | ==See Also== | ||
Revision as of 22:43, 1 February 2020
Problem
How many solutions does the equation tan
have on the interval ![$[0,2\pi]?$](http://latex.artofproblemsolving.com/e/c/c/ecccefb4c460857891d99273237d9deb9ecf41c1.png)
Solution
Draw a graph of tan
and cos
tan has a period of 
asymptotes at 
and zeroes at 
. It is positive from 
and negative elsewhere.
cos has a period of 
and zeroes at 
. It is positive from 
and negative elsewhere.
Drawing such a graph would get 
~lopkiloinm
Solution (Algebraically)
. Applying double angle identities for both, we have
![\[\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin x \cos x}{2\cos^{2}x-1}\]](http://latex.artofproblemsolving.com/8/c/a/8ca20f402a19e616d74fa3cb94b5e83b80f24e1c.png)
Applying half angle identities on the RHS, we have 
.
Setting both sides equal and squaring,
![\[\frac{2\sin x \cos x}{2\cos^{2}x-1}=\pm\sqrt{\frac{\cos x + 1}{2}}\]](http://latex.artofproblemsolving.com/7/9/7/79725325a7beaaeb12f5dfa7c1c9982d586e92da.png)
![\[\frac{4\sin^2 x \cos^2 x}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}\]](http://latex.artofproblemsolving.com/9/5/7/9578f78e922c7017a1e1c8b89728a5fc901b6fbc.png)
Since 
, we can substitute 
to convert the whole equation into cosine.
![\[\frac{4(1-\cos^2 x) (\cos^2 x)}{4\cos^{4}x-4\cos^2 x+1}=\frac{\cos x + 1}{2}\]](http://latex.artofproblemsolving.com/1/6/9/169ca5d24b88e0f04b0577500d669b2b9a914df8.png)
Cross multiplying, we get
![\[8(1-\cos^2 x) (\cos^2 x)=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)\]](http://latex.artofproblemsolving.com/3/2/f/32f12fcfddcc9e11b5e36fae97466118599d1ff5.png)
![\[0=(4\cos^{4}x-4\cos^2 x+1)(1+\cos x)-8(1-\cos^2 x) (\cos^2 x)\]](http://latex.artofproblemsolving.com/a/3/5/a351775bd4b1b706acaa7204c3330d070a155d25.png)
Without expanding anything, we can see that the first two polynomials will expand into a polynomial with degree 
and the 
term will expand into a polynomial with degree 
. This means that overall, the polynomial will have degree . From this, we can see that there are 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
