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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 10"

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We know <math>\cos2x=2\cos^2x-1=1-2\sin^2x</math>
 
We know <math>\cos2x=2\cos^2x-1=1-2\sin^2x</math>
  
−
So, <math>\left(\frac{1-\cos2x}2\right)^5+\left(\frac{1+\cos2x}2\right)^5=\frac{29}{16}\cos^42x</math>
+
So, <math>\sin^{10}x+\cos^{10}x=\left(\frac{1-\cos2x}2\right)^5+\left(\frac{1+\cos2x}2\right)^5</math> <math>=\frac{2(1+\binom52\cos^22x+\binom 54\cos^42x)}{32}</math>
  
−
On Simplification, <math>24\cos^42x-10\cos^22x-1=0\implies 24(\cos^22x)^2-10\cos^22x-1=0</math>
+
So, <math>\frac{1+10\cos^22x+5\cos^42x}{16}=\frac{29}{16}\cos^42x</math>
 +
 
 +
On Simplification, <math>24\cos^42x-10\cos^22x-1=0</math> which is a quadratic equation in <math>\cos^22x</math>
  
 
So, <math>\cos^22x=\frac{10\pm \sqrt{10^2-4\cdot24\cdot(-1)}}{2\cdot24}=\frac12</math> or <math>-\frac1{12}</math>
 
So, <math>\cos^22x=\frac{10\pm \sqrt{10^2-4\cdot24\cdot(-1)}}{2\cdot24}=\frac12</math> or <math>-\frac1{12}</math>
  
−
As <math>x</math> is real, <math>\cos^22x\ge 0\implies \cos^22x=\frac12\implies 2\cos^22x=1</math>
+
As <math>x</math> is real, <math>0\le\cos^22x\le1\implies \cos^22x=\frac12\implies 2\cos^22x=1</math>
  
 
Hence, <math>\cos4x=0\implies 4x=(2n+1)90</math> or <math>x=(2n+1)22.5</math> where <math>n</math> is any integer.
 
Hence, <math>\cos4x=0\implies 4x=(2n+1)90</math> or <math>x=(2n+1)22.5</math> where <math>n</math> is any integer.
  
−
So, <math>0\le \frac{(2n+1)90}4\le2007\implies -\frac12\le n\le \frac{882}{20}=44.1\implies 0\le n\le44</math>
+
So, <math>0\le \frac{(2n+1)90}4\le2007\implies -\frac12\le n\le \frac{882}{20}=44.1\implies 0\le n\le44</math>
  
 
==See Also==
 
==See Also==
 
{{Mock AIME box|year=2006-2007|n=2|num-b=9|num-a=11}}
 
{{Mock AIME box|year=2006-2007|n=2|num-b=9|num-a=11}}

Latest revision as of 23:09, 1 February 2013

Problem

Find the number of solutions, in degrees, to the equation $\sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,$ where $0^\circ \le x^\circ \le 2007^\circ.$

Solution

We know $\cos2x=2\cos^2x-1=1-2\sin^2x$

So, $\sin^{10}x+\cos^{10}x=\left(\frac{1-\cos2x}2\right)^5+\left(\frac{1+\cos2x}2\right)^5$ $=\frac{2(1+\binom52\cos^22x+\binom 54\cos^42x)}{32}$

So, $\frac{1+10\cos^22x+5\cos^42x}{16}=\frac{29}{16}\cos^42x$

On Simplification, $24\cos^42x-10\cos^22x-1=0$ which is a quadratic equation in $\cos^22x$

So, $\cos^22x=\frac{10\pm \sqrt{10^2-4\cdot24\cdot(-1)}}{2\cdot24}=\frac12$ or $-\frac1{12}$

As $x$ is real, $0\le\cos^22x\le1\implies \cos^22x=\frac12\implies 2\cos^22x=1$

Hence, $\cos4x=0\implies 4x=(2n+1)90$ or $x=(2n+1)22.5$ where $n$ is any integer.

So, $0\le \frac{(2n+1)90}4\le2007\implies -\frac12\le n\le \frac{882}{20}=44.1\implies 0\le n\le44$

See Also

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
Problem 9 		Followed by
Problem 11
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