Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Review Exercises: 49

Answer

There are three solutions in the interval $[0^{\circ},360^{\circ})$: $\theta = 70.5^{\circ}, 180^{\circ}, 289.5^{\circ}$

Work Step by Step

$3~cos^2~\theta+2~cos~\theta-1 = 0$ We can use the quadratic formula to find the solutions for $cos~\theta$: $cos~\theta = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $cos~\theta = \frac{-2 \pm \sqrt{2^2-(4)(3)(-1)}}{(2)(3)}$ $cos~\theta = \frac{-2 \pm \sqrt{16}}{6}$ $cos~\theta = \frac{1}{3}, -1$ When $cos~\theta = \frac{1}{3}$: $\theta = 70.5^{\circ}, 289.5^{\circ}$ When $cos~\theta = -1$: $\theta = 180^{\circ}$ There are three solutions in the interval $[0^{\circ},360^{\circ})$: $\theta = 70.5^{\circ}, 180^{\circ}, 289.5^{\circ}$
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