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Algebra Difficulty 5.0 AIME Find the answer

8. The solution set of the inequality 12tanx1>2\left|\frac{1}{\frac{2}{\tan x}-1}\right|>2 is

A number or a short expression. Spacing and $ signs are ignored.

Solution

8. xx(kπ+arctan43,kπ+arctan2)(kπ+arctan2,kπ+arctan4)kZ}\left.|x| x \in\left(k \pi+\arctan \frac{4}{3}, k \pi+\arctan 2\right) \cup(k \pi+\arctan 2, k \pi+\arctan 4) k \in Z\right\} 12tanx1>2\frac{1}{\frac{2}{\tan x}-1}>2 or 12tanx1<2\frac{1}{\frac{2}{\tan x}-1}<-2, and note that tanx0\tan x \neq 0.
We get tanx(43,2)(2,4)\tan x \in\left(\frac{4}{3}, 2\right) \cup(2,4)
Therefore, x(kπ+arctan43,kπ+arctan2)(kπ+arctan2,kπ+arctan4)kZx \in\left(k \pi+\arctan \frac{4}{3}, k \pi+\arctan 2\right) \cup(k \pi+\arctan 2, k \pi+\arctan 4) k \in \mathbf{Z}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.