solve for all possible x in [0,2pi):
(cscx+2)(cscx-√(2))=0
obviously either
cscx = -2 or cscx = √2
cscx < 0 in QIII,QIV
csc π/6 = 2, so
x = π+π/6 or 2π-π/6
x = 7π/6 or 11π/6
csc π/4 = √2, so
x = π/4 or 3π/4
So, the complete solution set is
π/4,3π/4,7π/6,11π/6