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: 55

Answer

The solution set is $$\{0^\circ+360^\circ n, 180^\circ+360^\circ n, n\in Z\}$$

Work Step by Step

$$\sin\theta\cos\theta-\sin\theta=0$$ 1) Solve the equation over the interval $[0^\circ,360^\circ)$ $$\sin\theta\cos\theta-\sin\theta=0$$ $$\sin\theta(\cos\theta-1)=0$$ $$\sin\theta=0\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=1$$ - For $\cos\theta=1$ Over the interval $[0^\circ, 360^\circ)$, there is only one value of $\theta$ where $\cos\theta=1$, which is $0^\circ$. - For $\sin\theta=0$ Over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\sin\theta=0$, which are $0^\circ$ and $180^\circ$. Therefore, overall, $$\theta=\{0^\circ, 180^\circ\}$$ (though $\theta=0^\circ$ appears in both cases, we only need to mention one time, since they are eventually the same point in the unit circle) 2) Solve the equation for all solutions - The integer multiples of the period of both the sine and cosine function is $360^\circ$. - We apply it to each solution found in step 1. So the solution set is $$\{0^\circ+360^\circ n, 180^\circ+360^\circ n, n\in Z\}$$
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.