Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 7 - Trigonometry and Periodic Functions - 7.8 Inverse Trigonometric Functions - Exercises and Problems for Section 7.8 - Exercises and Problems - Page 313: 10


No solution.

Work Step by Step

First, subtract $6$ from both sides to get $$2\cos \theta = 3$$ Divide both sides by $2$ to get $$\cos \theta = 3/2$$ However, $\cos \theta$ is bounded between $-1$ and $1$. $3/2$ is greater than $1$ ($2/2$) so there is no solution.
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