Question

15. DETAILS MY NOTES SWOKPRECALC13 6.2.051. Find the solutions of the equation that are in the interval $(0, 2\pi)$. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) $\tan^2 x \sin x = \sin x$ $x = $ Need Help?

Name: 15 details my notes swokprecalc13 62051 find the solutions of the equation that are in the interval 0 2pi enter your answers as a comma separated list if there is no solution enter no soluti 31683
Uploaded: 2022-12-19T18:30:35-08:00
Duration: 2 min 2 s
Channel: Manisha Sarker
Description: 15 details my notes swokprecalc13 62051 find the solutions of the equation that are in the interval 0 2pi enter your answers as a comma separated list if there is no solution enter no soluti 31683

          15. DETAILS
MY NOTES
SWOKPRECALC13 6.2.051.
Find the solutions of the equation that are in the interval $(0, 2\pi)$. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
$\tan^2 x \sin x = \sin x$
$x = $
Need Help?

15 details my notes swokprecalc13 62051 find the solutions of the equation that are in the interval 0 2pi enter your answers as a comma separated list if there is no solution enter no soluti 31683

Added by Sara L.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Manisha Sarker

12/19/2022

Step 1

We need to find the solutions in the interval $(0, 2\pi)$. First, move all terms to one side to set the equation to zero: $\tan^2 x \sin x - \sin x = 0$ Step 2: Factor out the common term, $\sin x$: $\sin x (\tan^2 x - 1) = 0$ Step 3: Set each factor equal to Show more…

Show all steps

Thanks for your feedback!

15. DETAILS MY NOTES SWOKPRECALC13 6.2.051. Find the solutions of the equation that are in the interval $(0, 2\pi)$. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.) $\tan^2 x \sin x = \sin x$ $x = $ Need Help?