Let #(2x+1)/(3x)=u#.
Then,
#sin(u)+sin(u)-3cos^2(u)=0#.
Using the trigonometric identity #sin^2(u)+cos^2(u)=1#,
#2sin(u)-3(1-sin^2(u))=0#,
#3sin^2(u)+2sin(u)-3=0#.
This is a quadratic equation in #sin(u)# it has roots given by the quadratic formulae of
#sin(u) = (-2pmsqrt(4+4*9))/(2*3)#,
#sin(u) = (-1pmsqrt(10))/(3)#.
Then, the general solution to #sin(u)=K# is #u=arcsin(K)+2npi# or #u = pi(1+2n) - arcsin(K)# where #n in Z#.
Notice that #sqrt(10)>3# so then #-1-sqrt(10)<-4/3#. So then the only solution set we need consider is #sin(u) = (-1+sqrt(10))/(3)#, as #abs(sin(x))<=1#.
Then, the solution set for #sin(u)# is
#u=arcsin((-1+sqrt(10))/3)+2npi# or #u=-arcsin((-1+sqrt(10))/3)+(2n+1)pi# where #n in Z#.
If #(2x+1)/(3x)=K# then,
#2x+1=3Kx#,
#x(2-3K)=-1#,
#x=1/(3K-2)#.
Then the solution set for #x# is
#x=1/(3arcsin((-1+sqrt(10))/3)+6npi-2)# or #x=1/(3(2n+1)pi-3arcsin((-1+sqrt(10))/3)-2)# where #n in Z#.
