Need Trig - Unit Circle help....please..

andipandi1220

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Feb 12, 2007
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I am supposed to find all the points (in terms of pi) on the unit circle for the equation:

2cot^2x+3cscx=0

I know that the equation can be changed to 2cot^2x+3(1+cot^2)=0
but then i am stuck...I need help figuring out how to get into an equation that can be factored. :?

Thanks!
 
Hello, andipandi1220!

A simple error . . .


Find all the points (in terms of pi) on the unit circle for the equation:
. . . 2cot2x+3cscx=0\displaystyle 2\cot^2x\,+\,3\csc x\:=\:0

I know that the equation can be changed to:
. . . 2cot2x+3(1+cot2x)=0  \displaystyle \,2\cot^2x\,+\,3(1\,+\,\cot^2x)\:=\:0\; . . . no

That identity is: csc2x=cot2x+1        cot2x=csc2x1\displaystyle \:\csc^2x \:=\:\cot^2x\,+\,1\;\;\Rightarrow\;\;\cot^2x\:=\:\csc^2x\,-\,1

The equation becomes: 2(csc2x1)+3cscx=0\displaystyle \:2(\csc^2x\,-\,1)\,+\,3\csc x\:=\:0

. . which simplifies to the quadratic: 2csc2x+3cscx2=0\displaystyle \:2\csc^2x\,+\,3\csc x\,-\,2\:=\:0

. . which factors: \(\displaystyle \:(\csc x\,+\,1)(2\csc x \,-\,1)\:=\:0\;\) . . . got it?

 
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