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{ | + | 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> |
− | + | So, <math>\frac{1+10\cos^22x+5\cos^42x}{16}=\frac{29}{16}\cos^42x</math> | |
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+ | 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, | + | 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, | + | 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 00:09, 2 February 2013
Problem
Find the number of solutions, in degrees, to the equation where
Solution
We know
So,
So,
On Simplification, which is a quadratic equation in
So, or
As is real,
Hence, or
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 |