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Mock AIME 1 2013 Problems/Problem 6: Difference between revisions

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Created page with "==Problem== Find the number of integer values <math>k</math> can have such that the equation <math>7\cos x+5\sin x=2k+1</math> has a solution. ==Solution== <math>2k+1</math> ..."
 
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Latest revision as of 15:33, 15 April 2015

Problem

Find the number of integer values $k$ can have such that the equation $7\cos x+5\sin x=2k+1$ has a solution.

Solution

$2k+1$ must be in the domain of $7\cos x+5\sin x$, so it suffices to find the maximum of this (the minimum is just $-1$ times the maximum). Taking the derivative and setting it to $0$, $\frac{d}{dx}(7\cos x+5\sin x)=-7\sin x+5\cos x=0$ Let $y=\cos x$. $5x=7\sqrt{1-x^2}, 25x^2=49(1-x^2), 74x^2=49, x=\frac{7}{\sqrt{74}}$. Therefore $7\cos x+5\sin x=\frac{49}{\sqrt{74}}+\frac{25}{\sqrt{74}}=\sqrt{74}$. Since $8<\sqrt{74}<9$, there are $\boxed{008}$ k's that work ($k=-4,-3,-2,-1,0,1,2,3$)

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