Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 10"
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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> | 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 | + | 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> |
Revision as of 13:48, 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,
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 |