Trigonometry (11th Edition) Clone

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

Chapter 6 - Quiz (Sections 6.1-6.3) - Page 282: 9


The solution set is $$\{\frac{\pi}{6}, \frac{2\pi}{3}\}$$

Work Step by Step

$$3\cot2x-\sqrt3=0$$ over the interval $[0,2\pi)$ We carry out algebraic operations here normally as any equations: $$\cot2x=\frac{\sqrt3}{3}$$ Now we can solve for $2x$. Over the interval $[0,2\pi)$, there are 2 values of $2x$ with which $\cot2x=\frac{\sqrt3}{3}$, which are $\frac{\pi}{3}$ and $\frac{4\pi}{3}$. Therefore, $$2x=\{\frac{\pi}{3},\frac{4\pi}{3}\}$$ That means, $$x=\{\frac{\pi}{6}, \frac{2\pi}{3}\}$$ This is the solution set of the problem.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.