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

[andipandi1220]

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!

 

[soroban]

Hello, andipandi1220!

A simple error . . .

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

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

That identity is: \(\displaystyle \:\csc^2x \:=\:\cot^2x\,+\,1\;\;\Rightarrow\;\;\cot^2x\:=\:\csc^2x\,-\,1\)

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

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

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

 

[andipandi1220]

yes!! so simple! :shock: thank you so much!