Stitz-Zeager_College_Algebra_e-book

Along with a which is given we nd ourselves in the

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Unformatted text preview: have f (0) = −1, for x = π we get f π = 1 and for x = π we get f (π ) = −1. 2 2 Since our original inequality is equivalent to f (x) ≤ 0, we are looking for where the function π is negative (−) or 0, and we get the intervals 0, π ∪ 56 , 2π . We can confirm our answer 6 graphically by seeing where the graph of y = 2 sin(x) crosses or is below the graph of y = 1. (−) 0 (+) 0 (−) 0 π 6 5π 6 2π y = 2 sin(x) and y = 1 2. We first rewrite sin(2x) > cos(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers, so we can advance to finding the zeros of f . Setting f (x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) cos(x) − cos(x) = 0 or cos(x)(2 sin(x) − 1) = 0. From cos(x) = 0, we get x = π + πk for integers k of which only x = π 2 2 π 1 and x = 32 lie in [0, 2π ). For 2 sin(x) − 1 = 0, we get sin(x) = 2 which gives x = π...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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