2008 iTest Problems/Problem 46

Problem

Let $S$ be the sum of all $x$ in the interval $[0,2\pi)$ that satisfy $\tan^2 x - 2\tan x\sin x=0$ . Compute $\lfloor10S\rfloor$ .

Solution

The equation can be factored into $(\tan x )(\tan x - 2\sin x) = 0$ . By the Zero Product Property, $\tan x = 0$ or $\tan x = 2\sin x$ .

If $\tan x = 0$ , then $x = 0$ or $x = \pi$ . If $\tan x = 2\sin x$ , then $\tfrac{1}{\cos x} = 2$ , so $\cos x = \tfrac12$ . That means $x = \tfrac{\pi}{3}$ or $x = \tfrac{5\pi}{3}$ .

Since $S$ is the sum of the solutions in the interval, $S = 0 + \pi + \tfrac{\pi}{3} + \tfrac{5 \pi}{3} = 3\pi$ , so $\lfloor10S\rfloor = \lfloor 30\pi \rfloor$ = \boxed{94}$.

2008 iTest (Problems)
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2008 iTest Problems/Problem 46

Problem

Solution

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