anonymous
  • anonymous
Something isn't right with the way I've done this question. see attached file 6 sin (x) = 1 + 9 sin (x) solve over domain 0 x 2pi
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
What attached file?
anonymous
  • anonymous
here it is
1 Attachment
anonymous
  • anonymous
1+9sin(x) -6sin(x) = 0 1 + 3sin(x) = 0 sin(x) = \(\sf \frac{1}{3}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
therefore, principal value of x = arcsin(1/3) = 19.47ยบ
anonymous
  • anonymous
negative or postive? Do I draw it in the first quadrant?
Michele_Laino
  • Michele_Laino
if we simplify that equation, we get: \(3 \sin x=-1\) therefore: \[\sin x = - \frac{1}{3}\]
Michele_Laino
  • Michele_Laino
again we have this drawing: |dw:1447353320315:dw|
anonymous
  • anonymous
When "solving" the equation, is it just the one angle in quad 4 that is the answer or both quad 3 and quad 4?
Michele_Laino
  • Michele_Laino
so your solutions are: \(x_1=199.47\) and \(x_2=340.53\)
Michele_Laino
  • Michele_Laino
we have to consider both angles in the third and fourth quadrant respectively
anonymous
  • anonymous
Ok, thank you again.. I have a few more I will ask.
Michele_Laino
  • Michele_Laino
ok! :)
anonymous
  • anonymous
just going to buy a few more owl bucks and be right back!

Looking for something else?

Not the answer you are looking for? Search for more explanations.