Trigonometry (11th Edition) Clone

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 49


The solution set is $$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, n\in Z\}$$

Work Step by Step

$$3\csc x-2\sqrt3=0$$ 1) Solve the equation over the interval $[0,2\pi)$ $$3\csc x-2\sqrt3=0$$ $$\csc x=\frac{2\sqrt3}{3}$$ Since $\sin x=\frac{1}{\csc x}$, this also equals to $$\sin x=\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}$$ Over the interval $[0,2\pi)$, there are two values of $x$ where $\csc x=\frac{2\sqrt3}{3}$ or equally $\sin x=\frac{\sqrt3}{2}$, which are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$. 2) Solve the equation for all solutions - The integer multiples of the period of the cosecant function is $2\pi$. - We apply it to each solution found in step 1. The results are the solution set. So the solution set is $$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, n\in Z\}$$
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.