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Anonymous
2016-01-21 01:50:24 Post No. 37477
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Anonymous
2016-01-21 01:50:24
Post No. 37477
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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?
