# 2000 Pan African MO Problems/Problem 1

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## Problem

Solve for $x \in R$: $$\sin^3{x}(1+\cot{x})+\cos^3{x}(1+\tan{x})=\cos{2x}$$

## Solution

Let $a = \sin{x}$ and $b = \cos{x}$. Substitution results in \begin{align*} a^3 (1 + \frac{b}{a}) + b^3 (1 + \frac{a}{b}) &= b^2 - a^2 \\ a^3 + a^2b + ab^2 + b^3 &= b^2 - a^2 \\ a^2 (a+b) + b^2 (a+b) &= (b+a)(b-a) \\ (a+b)(a^2 + b^2 + a - b) &= 0. \end{align*} By the Zero Product Property, $a = -b$ or $a^2 + b^2 + a - b = 0$. We can divide into two cases.

Case 1: $a = -b$

Substituting back results in $\sin{x} = -\cos{x}$. Since both sides don't equal each other when $\cos{x} = 0$, we can divide both sides by $\cos{x}$ to get $\tan{x} = -1$. Thus, $x = \frac{3 \pi}{4} + \pi n$, where $n$ is an integer.

Case 2: $a^2 + b^2 + a - b = 0$

Substituting back and applying a Pythagorean Identity results in $1 + \sin{x} - \cos{x} = 0$. Multiplying both sides by $\frac{\sqrt{2}}{2}$ and rearranging results in \begin{align*} \frac{\sqrt{2}}{2} &= \frac{\sqrt{2}}{2} \cos{x} - \frac{\sqrt{2}}{2} \sin{x} \\ \frac{\sqrt{2}}{2} &= \cos{(x + \frac{\pi}{4})} \end{align*} From the equation, $x = 0+ 2\pi n$ or $x = \frac{3 \pi}{2} + 2\pi n$, where $n$ is an integer. However, both are extraneous solutions because both make either $\tan{x}$ or $\cot{x}$ undefined, so there are no solutions in the case.

Therefore the solutions to the equation are $\boxed{x = \frac{3 \pi}{4} + \pi n, n \in \mathbb{Z}}$.