Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.1 - Solving Trigonometric Equations - 6.1 Problem Set - Page 326: 25

Answer

a) $= (2x -1)(x-3)$ b) $= (2cos\theta -1)(cos\theta - 3)$

Work Step by Step

a) $2x^{2} -7x + 3$ $= 2x^{2} - 6x - 1x + 3$ $= 2x(x - 3) - 1(x - 3)$ $= (2x -1)(x-3)$ b) $2(cos\theta)^{2} -7cos\theta+3$ Let $x = cos\theta$ to get $2x^{2} - 7x + 3$ (Exact same equation as in Section a)) $=(2x-1)(x-3)$. Substitute $cos\theta$ into $x$ to get $ (2cos\theta -1)(cos\theta - 3)$
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