Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 10"
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Find the number of solutions, in degrees, to the equation <math>\sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>0^\circ \le x^\circ \le 2007^\circ.</math> | Find the number of solutions, in degrees, to the equation <math>\sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>0^\circ \le x^\circ \le 2007^\circ.</math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | We know <math>\cos2x=2\cos^2x-1=1-2\sin^2x</math> |
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| + | So, <math>\left(\frac{1-\cos2x}2\right)^5+\left(\frac{1+\cos2x}2\right)^5=\frac{29}{16}\cos^42x</math> | ||
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| + | On Simplification, <math>24\cos^42x-10\cos^22x-1=0\implies 24(\cos^22x)^2-10\cos^22x-1=0</math> | ||
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| + | So, <math>\cos^22x=\frac{10\pm \sqrt{10^2-4\cdot24\cdot(-1)}}{2\cdot24}=\frac12</math> or <math>-\frac1{12}</math> | ||
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| + | As <math>x</math> is real, <math>\cos^22x\ge 0\implies \cos^22x=\frac12\implies 2\cos^22x=1</math> | ||
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| + | So, <math>\cos4x=0\implies 4x=(2n+1)90^\circ</math> where <math>n</math> is any integer. | ||
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| + | So, <math>0\le (2n+1)90\le2007\implies -\frac12\le n\le \frac{213}{20}<11\implies 0\le n\le10</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}} | ||
Revision as of 13:38, 31 January 2013
Problem
Find the number of solutions, in degrees, to the equation 
where 
Solution
We know 
So, 
On Simplification, 
So, 
or 
As 
is real, 
So, 
where 
is any integer.
So, 
See Also
| 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 | ||
