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Mathematics 14 Online
OpenStudy (anonymous):

how to solve the following equation: csc^2x-4=0

OpenStudy (mertsj):

Change csc^2x to 1/sin^2x

OpenStudy (mertsj):

\[\frac{1}{\sin ^2x}=4\]

OpenStudy (mertsj):

Now take reciprocals of both sides.

OpenStudy (mertsj):

\[\sin ^2x=\frac{1}{4}\]

OpenStudy (mertsj):

Now you can take it from there.

OpenStudy (anonymous):

Do I take the square root?

OpenStudy (mertsj):

yes

OpenStudy (anonymous):

so, x=pi/6 and 5pi/6

OpenStudy (mertsj):

no. sinx= + or - 1/2 so there is an answer in each quadrant.

OpenStudy (anonymous):

so, x=pi/6,5pi/6,7pi/6 and 11pi/6

OpenStudy (mertsj):

yes

OpenStudy (mertsj):

Unless you want all the solutions in which case it would be your 4 answers + 2pi n

OpenStudy (anonymous):

could you say :pi/6+npi,5pi/6+npi or pi/6+2npi,5pi/6+2npi

OpenStudy (anonymous):

as a general solution...

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