Stitz-Zeager_College_Algebra_e-book

8 use the cases and diagrams in the proof of the law

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Unformatted text preview: where they intersect. y = 3(sin(x))3 and y = (sin(x))2 y= 1 (cos(x))2 and y = tan(x) + 3 3. In the equation cos(2x) = 3 cos(x) − 2, we have the same circular function, namely cosine, on both sides but the arguments differ. Using the identity cos(2x) = 2 cos2 (x) − 1, we obtain a ‘quadratic in disguise’ and proceed as we have done in the past. cos(2x) 2 cos2 (x) − 1 2 cos2 (x) − 3 cos(x) + 1 2u2 − 3u + 1 (2u − 1)(u − 1) = = = = = 3 cos(x) − 2 3 cos(x) − 2 (Since cos(2x) = 2 cos2 (x) − 1.) 0 0 Let u = cos(x). 0 This gives u = 1 or u = 1. Since u = cos(x), we get cos(x) = 1 or cos(x) = 1. Solving 2 2 π cos(x) = 1 , we get x = π + 2πk or x = 53 + 2πk for integers k . From cos(x) = 1, we get 2 3 π x = 2πk for integers k . The answers which lie in [0, 2π ) are x = 0, π , and 53 . Graphing 3 y = cos(2x) and y = 3 cos(x) − 2, we find, after a little extra effort, that the curves intersect in three places on [0, 2π ), and the x-coordinates of these points con...
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