Let's solve the equation for #csc(x):#
#csc^2x-4=0#
#csc^2x=4#
#cscx=+-2#
Knowing that #cscx=1/sinx#,
#1/sinx=+-2#
Flip both sides:
#sinx=+-1/2#
First, let's consider the values of #x# which yield #1/2# for the sine function. This is true for #x=pi/6# which is in the first quadrant. However, it's also true for #x=(5pi)/6,# which is in the second quadrant, as sine is also positive in the second quadrant.
So, #x=pi/6, x=(5pi)/6# give #1/2# for #sinx.#
Now, let's consider which values give us #sinx=-1/2:#
#x=(7pi)/6# in the third quadrant fulfills this; however, so does #x=(11pi)/6.#
So, we have #x=pi/6, (5pi)/6, (7pi)/6, (11pi)/6.# We want to simplify our answer and get an answer for #x# which will cover ALL solutions, even those beyond the interval #[0, 2pi].# Such solutions certainly exist, as the sine function is periodic, meaning that there are infinitely many values for #x# in the interval of all real numbers which give us our desired values for sine.
So, let's look at the following values:
#x=pi/6#
#x=(5pi)/6#
#x=(7pi)/6#
#x=(11pi)/6#
And add #2npi# to all of them, where #n# is any integer. We obtain #2npi# from recognizing that values of sine repeat for every #2pi# units, meaning our answers must repeat every #2pi# units.
#x=pi/6+2npi#
#x=(5pi)/6+2npi#
#x=(7pi)/6+2npi#
#x=(11pi)/6+2npi#