Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Review Exercises - Page 291: 37


The solutions are $0.4636$ and $3.6052$.

Work Step by Step

Add $1$ to both sides of the equation to obtain: $2\tan{x} = 1$ Divide $2$ to both sides of the equation to obtain: $\dfrac{2\tan{x}}{2}=\dfrac{1}{2} \\\tan{x} = \frac{1}{2} \\x=\tan^{-1}{(\frac{1}{2})} \\x=0.463647609 \\x\approx 0.4636$ The period of the tangent function is $\pi$. This means that $0.4636+\pi\approx 3.6052$, which is still within the interval $[0, 2\pi)$, is also a solution. Therefore, the solutions are $0.4636$ and $3.6052$.
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