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

Revision as of 12:48, 31 January 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, $\left(\frac{1-\cos2x}2\right)^5+\left(\frac{1+\cos2x}2\right)^5=\frac{29}{16}\cos^42x$

On Simplification, $24\cos^42x-10\cos^22x-1=0\implies 24(\cos^22x)^2-10\cos^22x-1=0$

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

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

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

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

@@ Line 12: / Line 12: @@
 As <math>x</math> is real,  <math>\cos^22x\ge 0\implies \cos^22x=\frac12\implies 2\cos^22x=1</math>
-Hence, <math>\cos4x=0\implies 4x=(2n+1)90^\circ</math> or <math>x=(2n+1)22.5^\circ</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>

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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

Revision as of 12:48, 31 January 2013

Problem

Solution

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