let #u=csc theta# then #3 csc^2 theta-2csctheta=2# becomes #3u^2-2u=2#
then if we put everything on one side we have #3u^2-2u-2=0#. So let's use the quadratic formula to solve.
Now take #a=3,b=-2, and c=-2# and put it in to the
formula #x=(-b+-sqrt(b^2-4ac))/(2a)#
#u=(2+-sqrt((-2)^2-4(3)(-2)))/(2(3))#
#=(2+-sqrt(28))/6#
#=(2+- 2sqrt(7))/6#
#=(cancel2(1+- sqrt(7)))/(cancel(6)3)#
#csctheta=(1+- sqrt(7))/3#
#theta=csc^-1 ((1+- sqrt(7))/3)=sin^-1 (3/(1+- sqrt(7)))#
#theta=sin^-1 (3/(1+ sqrt(7))) or theta = cancel(sin^-1 (3/(1- sqrt(7)))=undef#
#theta=55.37370265+360n^@, or theta =180^@-55.37370265+360n^@#
#theta=55.37370265^@+360n^@, or theta=124.6262974^@+360n^@#
Now let's put in n values (integers) to determine our final answer
#n=0, theta=55.37^@,124.636^@#
Note that if we put in other n-values our answers will be outside of the domain so our solution set is
#S={55.37^@,124.636^@}#
