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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

\[3\sin x-5+\csc x=0\]

\[3\sin ^2x-5\sin x+1=0\]

But I simplified that to \[\sin ^{2}x-3\sin x+1\]

I don't see how you get that?

Give me a sec

Oh wait.. I was looking at something else. My bad.

Would I use the quadratic formula to find the roots of 3sin^2x−5sinx+1=0?

\[3sinx \times sinx-5\times sinx+\frac{1}{sinx}\times sinx=0 \times sinx\]

yes

\[\frac{ 3\pm \sqrt{5}}{ 2 }\] But then what? I can't isolate for x, can I?

sin x = (3+sqrt5)/2 or sin x = (3-sqrt5)/2

So would I find the arcsin of (3+sqrt5)/2 and (3-sqrt5)/2 for x?

Can't be the first one because sin can't be bigger than 1.

so sin x = .3820

Does there happen to be a way I could solve this without using a calculator?

If you have a table of values you could.

Did you get 22.5 degrees and 157.5 degrees?

Do we need my assistance here or are we ok?