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TRIG HELP: Not even a HW question - I'm...
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TRIG HELP: Not even a HW question - I'm beyond basic college Trig, in college Calc now. I was just doing some old Trig equations for practice and came upon a really tricky one I hadn't done before. Please help me find where I went wrong.
>Pro-Tip: You can't.

Solve for x on the interval [0degrees,360degrees):
sqrt(3)cos(x)-sqrt(3)=sin(x)
sqrt(3)(cos(x)-1)=sin(x)
sqrt(3)(cos(x)-1)-sin(x)=0
sqrt(3)-(sin(x)/(cos(x)-1))=0
sqrt(3)=sin(x)/(cos(x)-1)
sqrt(3)=(-1)(sin(x)/(1-cos(x)))
-sqrt(3)=sin(x)/(1-cos(x))
(-sqrt(3))(1-cos(x))=sin(x)
((-sqrt(3))(1-cos(x)))/(sin(x))=1
(1-cos(x))/(sin(x))=(-1/sqrt(3))
tan(x/2)=(-1/sqrt(3))
(x/2)=330degrees, 150degrees
x=660degrees, 300degrees

That's my answer, anyways. I think the book gives x=330degrees and 0degrees. Where did I go wrong in my algebra?
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Embarrasing
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>>37477
Just put it in a calculator and you will see that your solution is (partially correct):

http://www.wolframalpha.com/input/?i=sqrt%283%29cos%28300%29+-sqrt%283%29+-+sin%28300%29

But you divided by (cos x - 1) which is only allowed if cos x is not 1. You need to check the values of x with cos x = 1 and you will find that x = 0 is also a solution.

sqrt(3)cos(x)-sqrt(3)=sin(x)
sqrt(3)(cos(x) - 1) = sin(x)
sqrt(3)(cos(x) - 1)/sin(x) = 1 (sin x not 0)
(cos(x) - 1)/sin(x) = 1/sqrt(3)
(1 - cos(x))/sin(x) = -1/sqrt(3)
tan(x/2) = -1/sqrt(3)