Stitz-Zeager_College_Algebra_e-book

7 1031 82 a 88 b 92 c 1347 q 7859 2681

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Unformatted text preview: rm our results. 4. To solve cos(3x) = 2 − cos(x), we use the same technique as in the previous problem. From Example 10.4.3, number 4, we know that cos(3x) = 4 cos3 (x) − 3 cos(x). This transforms the equation into a polynomial in terms of cos(x). cos(3x) − 3 cos(x) 2 cos3 (x) − 2 cos(x) − 2 4u3 − 2u − 2 4 cos3 (x) = = = = 2 − cos(x) 2 − cos(x) 0 0 Let u = cos(x). 10.7 Trigonometric Equations and Inequalities 735 To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2 . We get either u − 1 = 0 or 4u2 − 2u − 2 = 0, and since the discriminant of the latter is negative, the only real solution to 4u3 − 2u − 2 = 0 is u = 1. Since u = cos(x), we get cos(x) = 1, so x = 2πk for integers k . The only solution which lies in [0, 2π ) is x = 0. Graphing y = cos(3x) and y = 2 − cos(x) on the same set of axes over [0, 2π ) shows that the graphs intersect at what appears to be (0, 1), as required. y = cos(2x) and y = 3 cos(x)...
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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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