I am not sure of your notation here. The use of the comma to me, signifies:
#tan3alpha-1=0 \ \ \ \ [1]#
and:
#tan3alpha-1=3 \ \ \ \ \ [2]#
#[1]#
#tan3alpha-1=0#
#tan3alpha=1#
#3alpha=arctan(tan3alpha)=arctan(1)#
#3alpha=arctan(1)# #-> arctan(1)=45^@, 225^@#
These angles lie in the I and III quadrants.
So we have:
#3alpha=45=>alpha=45/3=15^@#
#3alpha=225=>alpha=225/3=75^@#
#[2]#
#tan3alpha-1=3#
#tan3alpha=4#
#tan3alpha=4#
#3alpha=arctan(tan(3alpha))=arctan(4)#
#3alpha=arctan(4)# -> #arctan(4)~~76^@, 256^@#
#alpha=arctan(4)/3~~76/3=25.3bar3#
#alpha=arctan(4)/3~~256/3=85.3bar3#
We need:
#0<=alpha<=90^@#
#alpha=15^@,75^@,25.3bar3^@,85.3bar3^@#
Exact solutions:
#alpha=15^@,75^@,(arctan(4)/3)^@, (180+arctan(4)/3)^@#