# Mock AIME 2 2006-2007 Problems/Problem 10

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## Problem

Find the number of solutions, in degrees, to the equation $\sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,$ where $0^\circ \le x^\circ \le 2007^\circ.$

## Solution

We know $\cos2x=2\cos^2x-1=1-2\sin^2x$

So, $\sin^{10}x+\cos^{10}x=\left(\frac{1-\cos2x}2\right)^5+\left(\frac{1+\cos2x}2\right)^5$ $=\frac{2(1+\binom52\cos^22x+\binom 54\cos^42x)}{32}$

So, $\frac{1+10\cos^22x+5\cos^42x}{16}=\frac{29}{16}\cos^42x$

On Simplification, $24\cos^42x-10\cos^22x-1=0$ which is a quadratic equation in $\cos^22x$

So, $\cos^22x=\frac{10\pm \sqrt{10^2-4\cdot24\cdot(-1)}}{2\cdot24}=\frac12$ or $-\frac1{12}$

As $x$ is real, $0\le\cos^22x\le1\implies \cos^22x=\frac12\implies 2\cos^22x=1$

Hence, $\cos4x=0\implies 4x=(2n+1)90$ or $x=(2n+1)22.5$ where $n$ is any integer.

So, $0\le \frac{(2n+1)90}4\le2007\implies -\frac12\le n\le \frac{882}{20}=44.1\implies 0\le n\le44$