Start with the definition of a function #csc(x)#:
#csc(x)=1/sin(x)#
This immediately implies the domain this equation can have solutions to #x!=pi*N# (that's where #sin(x)=0#), where #N# - any integer number
Using the definition of #csc(x)#, our equation is transformed into
#9sin(x)=1/sin(x)#
Multiplying by #sin(x)# both sides of equation we get
#9sin^2(x)=1#
#sin^2(x)=1/9#
#sin(x)=+-1/3#
Solutions for #sin(x)=1/3#:
#x=arcsin(1/3)+2pi*N# and, since #sin(phi)=sin(pi-phi)#,
#x=pi - arcsin(1/3)+2pi*N#
Solutions for #sin(x)=-1/3#, considering #sin(x)# is an odd function, that is #sin(-x)=-sin(x)#, is the same as for #sin(-x)=1/3#:
#x=-arcsin(1/3)+2pi*N# and, since #sin(phi)=sin(pi-phi)#,
#x=pi + arcsin(1/3)+2pi*N#
We can combine all these solutions into one expression:
#x=+-arcsin(1/3)+pi*N#, where #N# - any integer number.
