Difference between revisions of "2006 AIME I Problems/Problem 12"

Revision as of 19:06, 4 July 2013

Problem

Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$ , where $x$ is measured in degrees and $100< x< 200.$

Solution

Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$ , we have $a^3 + b^3 = (a+b)^3 \Leftrightarrow ab(a+b) = 0$ . But $a+b = 2\cos 4x\cos x$ , so we require $\cos x = 0$ , $\cos 3x = 0$ , $\cos 4x = 0$ , or $\cos 5x = 0$ .

Hence the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = 906$ .

Revision as of 19:51, 17 January 2008 (view source) Cmac89 (talk \| contribs) (cos(5x) solutions were incorrect) ← Older edit		Revision as of 19:06, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
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	[[Category:Intermediate Trigonometry Problems]]		[[Category:Intermediate Trigonometry Problems]]
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Difference between revisions of "2006 AIME I Problems/Problem 12"

Revision as of 19:06, 4 July 2013

Problem

Solution

See also