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$| 3 ^ { \tan ( \pi x ) } - 3 ^ { 1 - \tan ( \pi x ) } | \geq 2$

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up vote 3 down vote accepted

Let $y=3^{\tan(\pi x)}$ so that $3^{1-\tan(\pi x)}=\frac{3^1}{3^{\tan(\pi x)}}=\frac{3}{y}$. Now, your inequality becomes $|y-\frac{3}{y}|\ge 2$.

This can be solved with the boundary algorithm (sometimes called the test-point method)—solve the corresponding equation, $|y-\frac{3}{y}|=2$, plot those solutions on a number line, plot any values of $y$ for which part(s) of the equation are undefined (e.g. $y=0$), and test a value in each resulting interval to see if that interval satisfies the inequality you're solving.

Once you've got a solution for $y$, go back and use that to solve for $x$.

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This is equivalent to $(y^2-3)^2\ge4y^2$ with $y=3^{\tan(\pi x)}$. The situation of $y^2$ with respect to the roots of the polynomial $(z-3)^2-4z=(z-9)(z-1)$ yields the sign of the polynomial. As a result, the inequality holds if $\tan(\pi x)\le0$ or $\tan(\pi x)\ge1$. That is, the fractional part of $x$ should not be $\frac12$ nor in $(0,\frac14)$.

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Thanks so much for the explaination! –  Shawn Mar 21 '11 at 8:43
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