# 2011 AIME I Problems/Problem 9

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_{24\sin x} (24\cos x)=\frac{3}{2}$. Find $24\cot^2 x$.

## Solution 1

We can rewrite the given expression as $$\sqrt{24^3\sin^3 x}=24\cos x$$ Square both sides and divide by $24^2$ to get $$24\sin ^3 x=\cos ^2 x$$ Rewrite $\cos ^2 x$ as $1-\sin ^2 x$ $$24\sin ^3 x=1-\sin ^2 x$$ $$24\sin ^3 x+\sin ^2 x - 1=0$$ Testing values using the rational root theorem gives $\sin x=\frac{1}{3}$ as a root, $\sin^{-1} \frac{1}{3}$ does fall in the first quadrant so it satisfies the interval. There are now two ways to finish this problem.

First way: Since $\sin x=\frac{1}{3}$, we have $$\sin ^2 x=\frac{1}{9}$$ Using the Pythagorean Identity gives us $\cos ^2 x=\frac{8}{9}$. Then we use the definition of $\cot ^2 x$ to compute our final answer. $24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}$.

Second way: Multiplying our old equation $24\sin ^3 x=\cos ^2 x$ by $\dfrac{24}{\sin^2x}$ gives $$576\sin x = 24\cot^2x$$ So, $24\cot^2x=576\sin x=576\cdot\frac{1}{3}=\boxed{192}$.

## Solution 2

Like Solution 1, we can rewrite the given expression as $$24\sin^3x=\cos^2x$$ Divide both sides by $\sin^3x$. $$24 = \cot^2x\csc x$$ Square both sides. $$576 = \cot^4x\csc^2x$$ Substitute the identity $\csc^2x = \cot^2x + 1$. $$576 = \cot^4x(\cot^2x + 1)$$ Let $a = \cot^2x$. Then $$576 = a^3 + a^2$$. Since $\sqrt[3]{576} \approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\cot^2x = 24a = 24 \cdot 8 = \boxed{192}$.

~IceMatrix