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^@}#