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2006 AIME I Problems/Problem 12

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

$\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$

$\cos^3 (4x-x)+ \cos^3 (4x+x) = 8 \cos^3 4x \cos^3 x$

Using the sum and difference formulas for the cosine function:

$( \cos 4x \cos x + \sin 4x \sin x )^3 + ( \cos 4x \cos x - \sin 4x \sin x )^3 = 8 \cos^3 4x \cos^3 x$

Expanding the expression:

$( \cos^3 4x \cos^3 x + 3 \cos^2 4x \cos^2 x \sin 4x \sin x + 3 \cos 4x \cos x \sin^2 4x \sin^2 x + \sin^3 4x \sin^3 x )$

$+$

$( \cos^3 4x \cos^3 x - 3 \cos^2 4x \cos^2 x \sin 4x \sin x + 3 \cos 4x \cos x \sin^2 4x \sin^2 x - \sin^3 4x \sin^3 x )$

$=$

$8 \cos^3 4x \cos^3 x$

Combining like terms:

$2 \cos^3 4x \cos^3 x + 6 \cos 4x \cos x \sin^2 4x \sin^2 x = 8 \cos^3 4x \cos^3 x$

$-6 \cos^3 4x \cos^3 x + 6 \cos 4x \cos x \sin^2 4x \sin^2 x = 0$

Factoring $-6 \cos 4x \cos x$:

$( -6 \cos 4x \cos x ) ( \cos^2 4x \cos^2 x - \sin^2 4x \sin^2 x )= 0$

Using the difference of squares factorization:

$( -6 \cos 4x \cos x ) ( \cos 4x \cos x + \sin 4x \sin x ) ( \cos 4x \cos x - \sin 4x \sin x )= 0$

Using the sum and difference formulas for cosine in reverse:

$( -6 \cos 4x \cos x ) (\cos (4x-x)) ( \cos (4x+x))= 0$

$-6 \cos 4x \cos x \cos 3x \cos 5x = 0$

Setting each non-constant factor equal to 0:

$\cos x = 0$

$x = 90, 270, ....$

$\cos 3x = 0$

$3x = 90, 270, 450, 630, ....$

$x = 30, 90, 150, 210, ....$

$\cos 4x = 0$

$4x = 90, 270, 450, 630, 810, ....$

$x = 22.5, 67.5, 112.5, 157.5, 202.5, ....$

$\cos 5x = 0$

$5x = 90, 270, 450, 630, 810, 990, 1170, ....$

$x = 18, 54, 90, 126, 162, 198, 234, ....$

So the sum of the values of $x$ where $100< x< 200$ is:

$112.5 + 126 + 150 + 157.5 + 162 + 198 = 906$

See also

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