Question

solve csc^2(x)+cot(x) -1 =0 on the interval (0,2pi)

Keep all answers exact where possible, otherwise round to two decimal places.

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Answer #1

We know that csc^2(x)=1+cot^2(x)

So,.     csc^2(x)+cot(x) -1 =0

            1+cot^2(x)+cot(x)-1=0

  Or        cot^2(x)+cot(x)=0

            cot(x)[cot(x)+1]=0

  If.    cot(x)=0.         X=pi/2 ,3pi/2

and if.     cot(x)+1=0

                   Cot(x)=-1.      X=3pi/4 ,7pi/4


answered by: anonymous
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