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Where 0<=x<360, Solve the equation

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\[3\sin x-5+\csc x=0\]
\[3\sin ^2x-5\sin x+1=0\]
But I simplified that to \[\sin ^{2}x-3\sin x+1\]

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Other answers:

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?
\[3sinx \times sinx-5\times sinx+\frac{1}{sinx}\times sinx=0 \times sinx\]
\[\frac{ 3\pm \sqrt{5}}{ 2 }\] 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?

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