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how to solve the following equation: csc^2x-4=0
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Change csc^2x to 1/sin^2x
\[\frac{1}{\sin ^2x}=4\]
Now take reciprocals of both sides.
\[\sin ^2x=\frac{1}{4}\]
Now you can take it from there.
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Do I take the square root?
yes
so, x=pi/6 and 5pi/6
no. sinx= + or - 1/2 so there is an answer in each quadrant.
so, x=pi/6,5pi/6,7pi/6 and 11pi/6
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yes
Unless you want all the solutions in which case it would be your 4 answers + 2pi n
could you say :pi/6+npi,5pi/6+npi or pi/6+2npi,5pi/6+2npi
as a general solution...
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