We will use : #sin^2x-sin^2y=sin(x+y)sin(x-y)#.
Now, given that, #sin^2ntheta-sin^2(n-1)theta=sin^2theta#.
#:. sin{ntheta+(n-1)theta}sin{ntheta-(n-1)theta}=sin^2theta#.
#:. sin(2ntheta-theta)sintheta-sin^2theta=0#.
#:. sintheta{sin(2ntheta-theta)-sintheta}=0#.
#:. sintheta=0, or, sin(2ntheta-theta)=sintheta#.
#:. theta=kpi, or, 2ntheta-theta=(-1)^ktheta+kpi, k in ZZ#.
Case 1 : If, in the eqn., #2ntheta-theta=(-1)^ktheta+kpi, k in ZZ" is even"#,
say, #k=2m," we have, "2ntheta-theta=(+1)theta+2mpi#.
#:. 2ntheta-2theta=2mpi, i.e., theta=(mpi)/(n-1), m in ZZ, n!=1.#
Case 2 : If, in the eqn., #2ntheta-theta=(-1)^ktheta+kpi, k in ZZ" is odd"#,
say, #k=2m+1#, then, #2ntheta-theta=(-1)theta+(2m+1)pi,#
#:. 2ntheta=(2m+1)pi, or, theta=((2m+1)pi)/(2n), m in ZZ, n!=0#.
Hence, the Solution Set is #{kpi}uu{(mpi)/(m-1), m!=1}uu{((2m+1)pi)/(2m), n!=0}; k,m in ZZ#.
