Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 10"
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== Problem == | == Problem == | ||
| − | Find the number of solutions, in degrees, to the equation <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> | |
| − | - | + | 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> | |
| − | + | 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> | |
| + | |||
| + | 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. | ||
| + | |||
| + | 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== | ||
| + | {{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 
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 | ||
