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 $\lfloor10S\rfloor = \lfloor 30\pi \rfloor = \boxed{94}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 45
Followed by:
Problem 47
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Invalid username
Login to AoPS